the reaction releases 1,066.3 kJ of heat.
The balanced chemical equation for the reaction between Fe2O3 and CO to produce Fe and CO2 is:
Fe2O3 + 3CO → 2Fe + 3CO2
According to the equation, 1 mole of Fe2O3 reacts with 3 moles of CO to produce 2 moles of Fe and 3 moles of CO2.
Since there is an excess of CO, we can assume that all of the Fe2O3 will react completely. Therefore, the number of moles of CO needed can be calculated as:
3.40 mol Fe2O3 × (3 mol CO / 1 mol Fe2O3) = 10.2 mol CO
So, 10.2 moles of CO are needed to react completely with 3.40 moles of Fe2O3.
The heat released by the reaction can be calculated using the standard enthalpy of formation (ΔHf°) values for the compounds involved in the reaction. The ΔHf° values for Fe2O3, CO, Fe, and CO2 are -824.2 kJ/mol, -110.5 kJ/mol, 0 kJ/mol, and -393.5 kJ/mol, respectively.
To calculate the heat released, we can use the following formula:
ΔH = ΣnΔHf°(products) - ΣnΔHf°(reactants)
where ΣnΔHf° is the sum of the standard enthalpies of formation for the products and reactants, and n is the stoichiometric coefficient.
Plugging in the values, we get:
ΔH = (2 mol × -393.5 kJ/mol) + (3 mol × 0 kJ/mol) - (1 mol × -824.2 kJ/mol) - (3 mol × -110.5 kJ/mol)
= -1,066.3 kJ
Therefore, the reaction releases 1,066.3 kJ of heat.
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what is the enthalpy of combustion of a compound if it's enthalpy of formation is -520 KJ/mol and if the total enthaply of formation of its products is -670 KJ/mol
Answer:
Thus, to yield the enthalpy of the combustion reaction, we then sum the enthalpies of formation of the products, weighted by their stoichiometric coefficients, and subtract the enthalpy of formation of ethanol (-277.69kJ/mol). The result is ethanol's standard enthalpy of combustion of -1234.79kJ/mol.
is it true that in a chemical reaction new types of atoms that are different from those of the reactants are produced to form new substances
No, it is not true that new types of atoms are produced in chemical reaction.
What is meant by chemical reaction?A chemical reaction is a process that results in the chemical conversion of one group of chemical compounds into another.
Chemical reactions involve the rearrangement of atoms to form new substances, but atoms themselves are not created or destroyed in the process. This is known as law of conservation of mass, which states that the total mass of reactants in a chemical reaction is equal to the total mass of products.
Therefore, the types of atoms present in the reactants are also present in the products, just in different combinations and arrangements.
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What happens to the surroundings during an endothermic reaction?
Answer: The surroundings will lower in temperature.
Explanation:
Endothermic reactions draw heat from the surroundings in order to occur. So the surroundings will feel cold, as the heat is being used as energy in the reaction.
what is the temperature of 1.2 moles of Helium gas at 1950mm Hg if it occupies 15,500 ml of volume?
The temperature of 1.2 moles of Helium gas at 1950 mm Hg if it occupies 15,500 ml of volume is 131°C.
We know we want to use the ideal gas law because we have the moles, pressure, and volume and we want to get the pressure: PV = nRT, where P = pressure, V = volume, n = number of moles, R = gas constant, and T = Kelvin temperature.
Here, the following values are given: 1950 mmHg of pressure, 15,500 mL (15.5 L), 1.2 mol of material, and R = 62.3638 (according to a table of gas constants). To the equation, include these:
Solve for temperature as follows: (1950) x (15.5 L) = (1.2) x (62.3638) x Temperature Temperature = 403.88 K.
We can convert this to degrees Celsius by subtracting 273 from the total:
3 sig figs = 403.88 - 273 = 130.88 - 131 (3 sig figs)
This causes the temperature to be 131°C.
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suppose a student conducted a titration of an unknown solution of the weak acid ch3cooh with 0.880 m naoh. first, the student diluted 15.0 ml of the ch3cooh solution with 85.0 ml of water in an erlenmeyer flask and added 2 drops of the indicator, phenolphthalein. then, 0.880 m naoh was titrated into the diluted ch3cooh solution until the color of the solution changed to pink and the end point of the titration was reached. at the end point, 6.80 ml of 0.880 naoh was added to the ch3cooh solution. calculate the concentration of the ch3cooh solution.
The concentration of the CH₃COOH solution is 0.0399 M.
The balanced chemical equation for the reaction between CH₃COOH and NaOH is:
CH₃COOH + NaOH → CH₃COONa + H₂O
From the equation, it can be seen that one mole of NaOH reacts with one mole of CH₃COOH. Therefore, the number of moles of NaOH used in the titration is:
0.880 mol/L × 0.00680 L = 0.00598 mol
Since the dilution did not affect the number of moles of CH₃COOH, the number of moles of CH₃COOH in the original solution is also 0.00598 mol.
The volume of the original solution used in the titration is:
15.0 mL/100.0 mL = 0.15
Therefore, the concentration of the CH3COOH solution is:
0.00598 mol/0.15 L = 0.0399 M
As a result, the CH₃COOH solution has a concentration of 0.0399 M.
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Rank the following bonds from highest polarity to the lowest:
1 = most polar ; 4 = least polar
N--Si [ Select ] ["1", "2", "3", "4"]
O--Cl [ Select ] ["1", "2", "3", "4"]
C--S [ Select ] ["1", "2", "3", "4"]
H--N [ Select ] ["1", "2", "3", "4"]
The order of bonds from highest polarity to the lowest is:1. O--Cl2. H--N3. C--S4. N--Si
Polarity can be explained as the extent to which electron density is unevenly distributed between atoms within a molecule. This can result in the molecule having a partial positive or negative charge.
The polarity of the bond is decided by the electronegativity difference between the two atoms involved in the bond. If the difference is greater, the bond will be more polar.
O--Cl bond: Oxygen is more electronegative than chlorine, so the bond is polar. Hence, O--Cl has the highest polarity.
H--N bond: The difference in electronegativity between hydrogen and nitrogen is not as great as that between oxygen and chlorine, but it is still significant. Hence, H--N has the second-highest polarity.
C--S bond: The difference in electronegativity between carbon and sulfur is less significant, making the bond less polar. Hence, C--S has the third-highest polarity.
N--Si bond: The difference in electronegativity between nitrogen and silicon is the least significant of all the bonds given. Thus, N--Si has the lowest polarity.
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45. Perchloric acid (HCIO) reacts with aqueous potassium carbonate, forming carbon dioxide gas and water.
Answer:
2 HClO4 + K2CO3 → CO2 + 2 KClO4 + H2O
Explanation:
The balanced chemical equation for the reaction between perchloric acid (HClO4) and aqueous potassium carbonate (K2CO3) can be written as follows:
2 HClO4 + K2CO3 → CO2 + 2 KClO4 + H2O
In this reaction, two molecules of perchloric acid react with one molecule of aqueous potassium carbonate to produce one molecule of carbon dioxide gas, two molecules of potassium perchlorate, and one molecule of water.
Note that this reaction is a double displacement reaction, also known as a metathesis reaction, where the cations and anions of two different compounds exchange places, forming two new compounds. In this case, the hydrogen cation (H+) and the potassium cation (K+) exchange places, while the perchlorate anion (ClO4-) and the carbonate anion (CO3^2-) exchange places.
8. a second-order reaction has a half-life of 18 s when the initial concentration of reactant is 0.71 m. calculate the rate constant for this reaction
The rates are constant for this reaction 0.0782 M⁻¹s⁻¹. Second-order reactions are those in which the total of the exponents in the appropriate rate law of the chemical reaction equals two.
Second-order reactions are chemical processes that depend on either the concentrations of two first-order reactants or the concentration of one second-order reactant, according to the rate law equations provided below. A chemical reaction's half-life is the length of time it takes for half of the reactant to move through the reaction.
Second-order kinetics can be used to explain a variety of significant biological processes, including the creation of double-stranded DNA from two complementary strands. The total of the exponents in the rate law is equal to two in a second-order reaction. In this section, the two most typical types of second-order reactions will be thoroughly covered.
The differential (derivative) rate equation and the integrated rate equation are used to explain how the rate of a second-order reaction varies with the concentration of reactants or products. The integrated rate equation demonstrates how the concentration of species varies over time, whereas the differential rate law demonstrates how the reaction's rate changes over time.
Second order reaction for calculating rate constant is
t1/2 = 1/KCAO
Half-Life period = 18 s
Initial Concentration of reactant = 0.71M
18s = 1/K(0.71M)
K = 0.0782 M⁻¹s⁻¹
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a student performs three titrations in order to standardize a naoh solutions. the results are: 0.105 m, 0.0990 m, 0.110 m. calculate the average and the standard deviation. are the results acceptable, or should she perform more titrations?
The average of the three titrations is 0.104 m.
The standard deviation is 0.0055
The average of the three titrations is as follows:
(0.105 + 0.0990 + 0.110) / 3 = 0.104 m.
To calculate the standard deviation, we first need to calculate the variance. The variance is the sum of the squared differences from the mean, divided by the number of measurements minus one. Using the formula for variance, we get:
Variance = [(0.105 - 0.104)² + (0.0990 - 0.104)² + (0.110 - 0.104)²] / (3 - 1)
= 3.05 x 10⁻⁵
The standard deviation is the square root of the variance, which is:
standard deviation = √(3.05 x 10⁻⁵) = 0.0055
The standard deviation is relatively small compared to the average, indicating that the results are precise. However, we cannot determine if the results are accurate without knowing the true value of the NaOH solution concentration. Therefore, we need to compare the average to the expected value or to a certified reference material.
If the average is within an acceptable range of the expected value or certified reference material, then the results are acceptable. Otherwise, more titrations should be performed to increase the precision and accuracy of the measurements. It is recommended to consult with a teacher or a supervisor to determine the appropriate number of titrations needed for the specific experiment.
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what type of sold materials are typicall hard, have high melting points and poor electrical conductivities
Ionic and covalent network solids are hard, have high melting points, and are poor electrical conductors.
Commonly, materials that are hard, have high liquefying focuses, and poor electrical conductivities are ionic solids or covalent organization solids.
Ionic solids are made out of a three-layered exhibit of emphatically and adversely charged particles kept intact by electrostatic powers. The solid ionic connections between the particles make these solids hard and high softening, while the shortfall of free electrons makes them unfortunate channels of power.
Covalent organization solids, then again, are made out of an immense organization of covalent connections between particles in a gem cross section structure. These covalent bonds are major areas of strength for incredibly, these solids high softening focuses and hardness. Once more, the shortfall of free electrons implies these materials are unfortunate conduits of power.
Instances of ionic solids incorporate NaCl (table salt) and MgO (magnesium oxide), while instances of covalent organization solids incorporate jewel and silicon dioxide (quartz).
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To prepare 100.0 mL of 0.0300 M HCl, what volume of 0.500 M stock solution is required?
6.00 mL of the 0.500 M HCl stock solution is required to prepare 100.0 mL of 0.0300 M HCl.
We can use the following formula to calculate the volume of stock solution needed:
V(stock) x M(stock) = V(final) x M(final)
where V(stock) is the volume of the stock solution, M(stock) is the concentration of the stock solution, V(final) is the final volume of the diluted solution, and M(final) is the desired concentration of the diluted solution.
Plugging in the given values, we get:
V(stock) x 0.500 M = 100.0 mL x 0.0300 M
Solving for V(stock), we get:
V(stock) = (100.0 mL x 0.0300 M) / 0.500 M = 6.00 mL
A diluted solution is a solution that has been made weaker by adding more solvent (usually water) to a more concentrated solution. This results in a decrease in the concentration of the solute in the solution.
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Using the "NAS" idea, how many electrons are "needed" for the compound PBr3?
A. 12
B. 32
C. 26
D. 21
Answer:
32
Explanation:
aluminum is produced commercially by the electrolysis of al2o3 in the presence of a molten salt. if a plant has a continuous capacity of 1.15 million amp, what mass of aluminum can be produced in 2.40 h?
The mass of aluminum that can be produced in 2.40 h is 1390 kg aluminum production by Electrolysis.
The amount of aluminum produced can be calculated using Faraday's law, which states that the amount of substance produced is directly proportional to the amount of electric charge passed through the cell, and the molar mass of the substance.
The balanced chemical equation for the electrolysis of [tex]Al_2O_3[/tex] is:
[tex]2 Al_2O_3(l)[/tex] -> [tex]4 Al(l) + 3 O_2(g)[/tex]
From the equation, we see that for every 4 moles of aluminum produced, 6 moles of electrons are needed.
The charge passed through the cell can be calculated using the formula:
charge = current × time
Where current is measured in amperes (A) and time is measured in hours (h). We need to convert 2.40 h to seconds:
[tex]2.40* 3600 = 8640 s[/tex]
The charge passed through the cell is:
charge = 1.15 million A × 8640 s = [tex]9.936 * 10^9[/tex] C
The number of moles of electrons passed through the cell can be calculated as:
moles of electrons = charge / Faraday's constant
Where Faraday's constant is 96,485 C/mol.
moles of electrons = [tex]\frac{9.936 * 10^9}{96,485} = 103,068[/tex] mol
Therefore, the number of moles of aluminum produced is half of the number of moles of electrons, or:
moles of Al = 0.5 × (103,068 mol) = 51,534 mol
The mass of aluminum produced can be calculated using the molar mass of aluminum, which is 26.98 g/mol:
mass of Al = moles of Al × molar mass of Al
mass of Al = [tex]51,534 * 26.98 = 1.39 * 10^6[/tex] g or 1390 kg
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pickles are made by immersing cucumbers in a concentrated saltwater solution. explain what happens to the cucumber in this process to cause it to shrink and taste salty.
When cucumbers are immersed in a concentrated saltwater solution, the process of osmosis occurs. Osmosis is the movement of water molecules across a selectively permeable membrane from an area of high water concentration to an area of low water concentration.
In this case, the concentrated saltwater solution outside the cucumber has a much lower water concentration than inside the cucumber. As a result, water from inside the cucumber moves out of the cell membrane and into the saltwater solution, causing the cucumber to shrink.
Additionally, the salt ions from the solution enter the cucumber through the cell membrane, which is permeable to ions. The presence of salt ions in the cucumber affects the taste, making it salty. The salt also acts as a preservative, inhibiting the growth of microorganisms that can spoil the cucumber. This is why pickles have a longer shelf life than fresh cucumbers.
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the second-order decomposition of no2 has a rate constant of 0.255 m-1s-1. how much no2 decomposes in 8.00 s if the initial concentration of no2 (1.00 l volume) is 1.33 m? the second-order decomposition of no2 has a rate constant of 0.255 m-1s-1. how much no2 decomposes in 8.00 s if the initial concentration of no2 (1.00 l volume) is 1.33 m? 0.85 mol 0.97 mol 1.66 mol 1.9 mol 0.36 mol
In 8.00 seconds, approximately 0.84 mol of NO2 decomposes. The closest answer among the given options is 0.85 mol.
To find out how much NO2 decomposes in 8.00 s given the initial concentration and rate constant, we can use the second-order reaction formula:
1/[A]t = kt + 1/[A]₀
Where:
- [A]t is the concentration of NO2 at time t (which we want to find)
- k is the rate constant (0.255 M⁻¹s⁻¹)
- t is the time (8.00 s)
- [A]₀ is the initial concentration of NO2 (1.33 M)
Step 1: Plug the values into the equation.
1/[A]t = (0.255 M⁻¹s⁻¹)(8.00 s) + 1/(1.33 M)
Step 2: Calculate the value on the right side of the equation.
1/[A]t = 2.04 M⁻¹
Step 3: Solve for [A]t (concentration of NO2 at 8.00 s).
[A]t = 1/2.04 M⁻¹ = 0.49 M
Step 4: Calculate the change in concentration (how much NO2 decomposes).
Change in concentration = [A]₀ - [A]t = 1.33 M - 0.49 M = 0.84 M
Step 5: Convert the change in concentration to moles (since the volume is 1.00 L, the change in concentration is equal to the change in moles).
Change in moles = 0.84 mol
So, the correct option is 0.85 as it is closest to the answer.
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a. it is impossible to anticipate the ph profile of an acid and base without performing the experiment in the lab. b. the ph profile is expected to behave similarly to the example of hydrochloric acid and sodium hydroxde provided in the lab manual. c. the ph profile is expected to behave similarly to the example of acetic acid and sodium hydroxide provided in the lab manual. d. the identity of the acid does not play a role in determining the ph profile of the reaction between an acid and a base.
Answer:
a. it is impossible to anticipate the pH profile of an acid and base without performing the experiment in the lab.
Explanation:
PLEASE HELPPP!! I'M STUCK ON THISS
The arrangement of radial, symmetric, and asymmetric letters is found in the attachment.
What are radial, symmetric, and asymmetric letters?Bilateral Letters: These are letters that have a symmetrical shape where the left and right sides are mirror images of each other. In other words, if you were to draw a vertical line down the center of the letter, both sides would be identical.
Examples of bilateral letters include B, C, D, E, G, H, K, M, O, P, Q, R, S, U, and V.
Radial Letters: These are letters that have a symmetrical or circular shape around a central point. If you were to draw a circle around the letter, it would fit within that circle.
Examples of radial letters include A, C, D, M, and O.
Asymmetric Letters: These are letters that do not have symmetry or balance. If you were to draw a vertical line down the center of the letter, the two sides would not be mirror images of each other.
Examples of asymmetric letters include I, J, L, N, T, U, V, W, X, Y, and Z.
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Suppose only 5,550J heat was used to warm up the same 55.0g of water. If the water started out at 25 degrees celsius what would its final temperature become?
Answer:
What is the heat in joules required to raise the temperature of 25 grams of water from 0 degrees C to 100 degrees C? What is the heat in calories? Answer: 10450 J or 2500 calories of heat energy are required to raise the temperature of 25 grams of water from 0 degrees C to 100 degrees C.
according to the galvanic cell video provided in the introduction page of this module, which statement is true about the four main parts of the galvanic cell? group of answer choices the anode is the electrode that receives electrons. an external circuit with a multimeter will be used to transfer electrons. there are two electrodes and they will be immersed in the same metal solutions. the cathode is the electrode where oxidation occurs. salt bridge will only be placed in one of the cells.
According to galvanic cell video which is provided in the introduction page of this module, the statement which is true about the four main parts of the galvanic cell will be; "The cathode will be the electrode where reduction can occurs." Option D is correct.
A galvanic cell, also termed as a voltaic cell, in an electrochemical cell which uses a spontaneous redox reaction to generate an electric current. The cell consists of two half-cells, each containing an electrode and an electrolyte solution. The two half-cells are connected by a wire as well as a salt bridge.
The other statements are not correct; The anode is the electrode where oxidation occurs, not where it receives electrons.
An external circuit with a voltmeter (not multimeter) is used to measure the potential difference (voltage) between the two electrodes.
There are two electrodes, but they are immersed in different solutions (not the same metal solutions).
The salt bridge is placed in both cells to allow the flow of ions to maintain electrical neutrality, not just in one cell.
Hence, D. is the correct option.
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--The given question is incomplete, the complete question is
"According to the galvanic cell video provided in the introduction page of this module, which statement is true about the four main parts of the galvanic cell? group of answer choices A) the anode is the electrode that receives electrons. B) an external circuit with a multimeter will be used to transfer electrons. C) there are two electrodes and they will be immersed in the same metal solutions. D) the cathode is the electrode where oxidation occurs. E) salt bridge will only be placed in one of the cells."--
Calculate the Heat of Formation for the following reaction.
2NO + O2 ---> 2NO2
The heat of formation of the given reaction is -56.4 kJ/mol.
What is Heat?
Heat is a form of energy that flows from a hotter object to a colder object. It is a type of energy transfer that occurs due to a temperature difference between two objects. The direction of heat flow is always from the object with higher temperature to the object with lower temperature until they reach thermal equilibrium.
The heat of formation of a compound is defined as the enthalpy change when one mole of the compound is formed from its constituent elements in their standard states at a given temperature and pressure. The standard state of an element is its most stable form at 1 atm pressure and a specified temperature.
Using the heat of formation values from standard tables, we can calculate the heat of formation of the products and reactants in the given reaction as follows:
Heat of formation of NO2 = -33.2 kJ/mol
Heat of formation of NO = +90.3 kJ/mol (since NO is an unstable gas at room temperature, we use the heat of formation of NO at 298 K instead of the standard heat of formation)
Heat of formation of O2 = 0 kJ/mol
Therefore, the heat of formation of the given reaction is:
ΔHf = Σ(heat of formation of products) - Σ(heat of formation of reactants)
ΔHf = [2(-33.2 kJ/mol)] - [2(90.3 kJ/mol) + 0 kJ/mol]
ΔHf = -56.4 kJ/mol
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the haber process synthesizes ammonia at elevated temperatures and pressures from nitrogen gas and hydrogen gas. what volume of ammonia would be produced if 350 l of nitrogen gas and 900 l of hydrogen gas is combined at stp? assume the reaction goes to completion.
When 350 L of nitrogen gas and 900 L of hydrogen gas are combined at STP using the Haber process, 329 L of ammonia gas is produced.
Assuming the reaction goes to completion, the Haber process synthesizes ammonia at elevated temperatures and pressures from nitrogen gas and hydrogen gas. As a result, balanced chemical equation can be given as:N2(g) + 3H2(g) → 2NH3(g)The balanced equation establishes that one mole of nitrogen reacts with three moles of hydrogen to produce two moles of ammonia.
This implies that the stoichiometric ratios of the reactants are 1:3. Thus, if 350 L of nitrogen and 900 L of hydrogen are mixed in a reaction vessel and the reaction proceeds to completion, the limiting reactant will be nitrogen. Hence, the number of moles of nitrogen present is calculated using the ideal gas equation n = PV/RT as follows:n(N2) = PV/RT = (1 atm) x (350 L) / (0.08206 L atm/mol K) x (273 K) = 14.07 mol
Similarly, the number of moles of hydrogen is calculated as:n(H2) = PV/RT = (1 atm) x (900 L) / (0.08206 L atm/mol K) x (273 K) = 36.25 molSince nitrogen is the limiting reactant, it will completely react with 1/3 of the amount of hydrogen present to produce ammonia. As a result, the amount of ammonia generated will be equivalent to the quantity of nitrogen that reacted. Therefore, the volume of ammonia is calculated as follows:n(NH3) = n(N2) = 14.07 molV(NH3) = n(NH3) x (RT/P) = 14.07 x (0.08206 x 273 / 1) / (1) = 329 L
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Determine the number of moles (n) for 22.5 g of carbon dioxide with a molecular formula of CO₂
There are 0.511 moles of CO 2 in 22.5 gm of the substance
Determine the number of moles of carbon dioxide, we need to use the molecular weight of CO₂ and the given mass of 22.5 g.
The molecular weight of CO₂ is:
c = 12.01 gm/mol
o = 16.00 gm/mol
2 x o = 2 x 16.00 gm/mol = 32.00 gm/mol
So, the molecular weight of CO₂ is 12.01 gm/mol + 32.00 gm/mol = 44.01 gm/mol.
Now, we can use the formula:
no of moles = Given mass / Molar mass
where n is the number of moles, m is the given mass, and M is the molecular weight.
Plugging in the values, we get:
n = 22.5 g / 44.01 g/mol
n = 0.511 moles"
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a buffer is made by combining 20.0 ml 0.250 m nh4cl with 30.0 ml 0.250 m nh3. a. calculate the ph of the buffer. 14 b. calculate the ph of the buffer after 1.00 ml of 6.00 m hcl is added and equilibrium is re- established. c. calculate the ph of the buffer after 1.00 ml of 6.00 m naoh is added to a fresh sample of the buffer and equilibrium is re-established.
a. The pH of the buffer is 9.24.
b. The pH of the buffer after the addition of HCl is 8.68.
c. The pH of the buffer after the addition of NaOH is 9.37.
The chemical equation for the reaction between NH₄⁺ and NH₃ is,
NH₄⁺ + NH₃ ⇌ NH₃ + NH₄⁺.
At equilibrium, the concentration of NH₄⁺ equals the concentration of NH₃. The Ka for NH₄⁺ is 5.6 × 10^-10.
Using the Henderson-Hasselbalch equation, pH = pKa + log([A⁻]/[HA]), where A⁻ is NH₃ and HA is NH₄⁺, we can calculate the pH of the buffer as follows:
pH = pKa + log([A⁻]/[HA])
pH = pKa + log([NH₃]/[NH₄⁺])
pH = pKa + log([0.250 M]/[0.250 M])
pH = -log(5.6 × 10^-10) + log(1)
pH = 9.24
When 1.00 mL of 6.00 M HCl is added to the buffer solution, it reacts with NH3 to form NH4+ and Cl^- ions.
The new concentration of NH₄⁺ is
[NH₄⁺] = 0.250 M + (1.00 × 10^-3 L)(6.00 M)/(20.0 mL + 1.00 mL) = 0.302 M.
The new concentration of NH₃ is [NH₃] = 0.250 M - (1.00 × 10^-3 L)(6.00 M)/(30.0 mL + 1.00 mL) = 0.196 M. Using the Henderson-Hasselbalch equation as before, we can calculate the new pH of the buffer as follows:
pH = pKa + log([A⁻]/[HA])
pH = pKa + log([NH₃]/[NH₄⁺])
pH = -log(5.6 × 10^-10) + log(0.196/0.302)
pH = 8.68
When 1.00 mL of 6.00 M NaOH is added to a fresh sample of the buffer, it reacts with NH₄⁺ to form NH₃ and Na+.
The new concentration of NH₄⁺ is,
[NH₄⁺] = 0.250 M - (1.00 × 10^-3 L)(6.00 M)/(20.0 mL + 1.00 mL)
= 0.198 M.
The new concentration of NH₃ is [NH₃] = 0.250 M + (1.00 × 10^-3 L)(6.00 M)/(30.0 mL + 1.00 mL) = 0.304 M. Using the Henderson-Hasselbalch equation as before, we can calculate the new pH of the buffer as follows:
pH = pKa + log([A⁻]/[HA])
pH = pKa + log([NH₃]/[NH₄⁺])
pH = -log(5.6 × 10^-10) + log(0.304/0.198)
pH = 9.37
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Find the mass, in grams, of 11.2 L H2.
The mass of 11.2 L of H2 gas at STP is 1.008 grams.
To find the mass of hydrogen gas in 11.2 L, we need to use the ideal gas law, which relates the pressure (P), volume (V), number of moles (n), and temperature (T) of a gas:
PV = nRT
where R is the gas constant.
To solve for the mass of H2, we need to know the pressure, temperature, and number of moles of the gas. Let's assume that the H2 is at standard temperature and pressure (STP), which is 0 degrees Celsius (273.15 K) and 1 atmosphere of pressure (101.3 kPa).
At STP, 1 mole of any gas occupies 22.4 L of volume. Therefore, the number of moles of H2 in 11.2 L can be calculated as:
n = (11.2 L) / (22.4 L/mol) = 0.5 mol
Now we can use the molar mass of hydrogen (2.016 g/mol) to convert the number of moles to mass:
mass = n x molar mass = 0.5 mol x 2.016 g/mol = 1.008 g
Therefore, the mass of 11.2 L of H2 gas at STP is 1.008 grams.
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calcium carbonate, also known as limestone, can be heated to produce calcium oxide (lime), an industrial chemical with a wide variety of uses, and carbon dioxide. how many grams of calcium carbonate must be decomposed to produce 5.00 l of carbon dioxide gas at stp?
19.7 g of calcium carbonate must be decomposed to produce 5.00 L of carbon dioxide gas at STP.
Calcium carbonate, also known as limestone, can be heated to produce calcium oxide (lime), an industrial chemical with a wide variety of uses, and carbon dioxide. In order to produce 5.00 L of carbon dioxide gas at STP, we need to determine how many grams of calcium carbonate must be decomposed.The chemical equation for the thermal decomposition of calcium carbonate is:CaCO3 (s) → CaO (s) + CO2 (g).To find out the number of grams of calcium carbonate that must be decomposed to produce 5.00 L of carbon dioxide gas at STP, we can use the following steps:
Step 1: Write down the given information.V = 5.00 L (volume of carbon dioxide gas)T = 273 K (temperature at STP)P = 1 atm (pressure at STP)
Step 2: Use the ideal gas law to calculate the number of moles of carbon dioxide gas.n = PV/RT = (1 atm) x (5.00 L) / (0.0821 L atm/mol K x 273 K) = 0.197 mol
Step 3: Use the stoichiometry of the chemical equation to calculate the number of moles of calcium carbonate required to produce this amount of carbon dioxide gas.
The stoichiometry of the chemical equation tells us that 1 mole of CaCO3 produces 1 mole of CO2. Therefore, we need 0.197 moles of CaCO3 to produce 0.197 moles of CO2.Step 4: Calculate the mass of calcium carbonate required using its molar mass.The molar mass of CaCO3 is 100.1 g/mol. Therefore, the mass of CaCO3 required is:m = n x M = 0.197 mol x 100.1 g/mol = 19.7 g.
Therefore, 19.7g calcium carbonate must be decomposed to produce 5l of co2 gas at STP .
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2.98x10^7+3.12x10^7 and a expressed answer in scientific notation
When adding numbers in scientific notation, we need to ensure that the exponents of 10 are the same.
2.98x10^7 + 3.12x10^7 can be rewritten as:
(2.98 + 3.12) x 10^7
= 6.10 x 10^7
Therefore, the sum of 2.98x10^7 and 3.12x10^7 in scientific notation is 6.10x10^7.
what mass of oxygen would form from 5 moles of water?
Answer:
Explanation: The equation of water is H2+1/2O2=H2O
->5 moles H2O*1 mol O2/2 mole
-> H2O=2.5 moles O2
->Now, molar mass of O2= 2*16=32gms
To find mass of O2,
-> No. of moles*Molar mass of O2
->2.5*32=0 gms
So, the mass of oxygen will be 80 gms.
what is the molecular geometry of the following and would you expect it to have a dipole moment? group of answer choices square planar, no octahedral, yes tetrahedral, yes octahedral, no square planar, yes
The molecular geometry of a given element is octahedral and it has no dipole moment. Therefore octahedral, No would be the correct answer.
The SF6 molecule has no dipole moment because each S−F bond dipole is balanced by one of equal magnitude pointing in the opposite direction of the other side of the molecule.
The three-dimensional configuration of the atoms that make up a molecule is known as molecular geometry. In addition to providing details about the molecule's overall shape, it also provides data on the bond lengths, bond angles, torsional angles, and any other geometrical factors that affect each atom's position.
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You are given the mass of one of the reactants in a chemical reaction and asked to find the mass of one of the products. What is the first step? Question 3 options: check to make sure the equation is balanced make the switch using the mole ratio give up get out the balance
Answer:
Check to make sure the equation is balanced
Explanation:
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ernest rutherford's experiments in which alpha rays pass through a thin piece of metal foil demonstrated that group of answer choices neutrons exist. the atom contains a tiny nucleus containing most of the atom's mass. light is made of particles. electrons exist. light is a wave.
Ernest Rutherford's experiments in which alpha rays pass through a thin piece of metal foil demonstrated that atom contains a tiny nucleus containing most of the atom's mass. The correct answer choice is "the atom contains a tiny nucleus containing most of the atom's mass."
The alpha-particle scattering experiment was performed by Ernest Rutherford in 1911. It was conducted to discover the nature of atomic structure. The experiment demonstrated that most of the mass of an atom and all of its positive charge are contained in a small nucleus at the center of the atom.
The electrons were found to occupy almost all of the remaining space. Therefore, the correct answer is the atom contains a tiny nucleus containing most of the atom's mass.
Therefore "the atom contains a tiny nucleus containing most of the atom's mass." is the correct answer.
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