Answer:
i. 1 : 2
ii. NaOH is the limiting reagent
iii. 292.5g
iv. 59.69%
Explanation:
I have the detailed and self-explanatory workings. will snap and post later. Battery percent at 15%, flashlight not working.
1. The ratio of CuSO₄ to NaOH is 1 : 2
2. The limiting reactant is NaOH.
3. The mass of Cu(OH)₂ that will precipitate out is 292.5 g.
4. The percentage yield of Cu(OH)₂ is 59.7%
1. Determination of the ratio of CuSO₄ to NaOH.
CuSO₄ + 2NaOH —> Cu(OH)₂ + Na₂SO₄
From the balanced equation above, we can see that the ratio of CuSO₄ to NaOH is 1 : 2
2. Determination of the limiting reactant.
CuSO₄ + 2NaOH —> Cu(OH)₂ + Na₂SO₄
Molar mass of CuSO₄ = 63.5 + 32 + (16×4) = 159.5 g/mol
Mass of CuSO₄ from the balanced equation = 1 × 159.5 = 159.5 g
Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol
Mass of NaOH from the balanced equation = 2 × 40 = 80 g
From the balanced equation above,
159.5 g of CuSO₄ reacted with 80 g of NaOH.
Therefore,
638.44 g of CuSO₄ will react with = (638.44 × 80) / 159.5 = 320.22 g of NaOH.
From the calculations made above, we can see that a higher mass (i.e 320.22 g) of NaOH than what was given (i.e 240 g) is needed to react completely with 638.44 g of CuSO₄.
Therefore, NaOH is the limiting reactant.
3. Determination of the mass of Cu(OH)₂ that will precipitate out.
CuSO₄ + 2NaOH —> Cu(OH)₂ + Na₂SO₄
Molar mass of Cu(OH)₂ = 63.5 + 2(16 + 1) = 97.5 g/mol
Mass of Cu(OH)₂ from the balanced equation = 1 × 97.5 = 97.5 g
From the balanced equation above,
80 g of NaOH reacted to produce 97.5 g of Cu(OH)₂.
Therefore,
240 g of NaOH will react to produce = (240 × 97.5) / 80 = 292.5 g of Cu(OH)₂
Thus, 292.5 g of Cu(OH)₂ precipitated out of the reaction.
4. Determination of the percentage yield.
Actual yield of Cu(OH)₂ = 174.6 gTheoretical yield of Cu(OH)₂ = 292.5 gPercentage yield of Cu(OH)₂ =?Percentage yield = (Actual /Theoretical) × 100
Percentage yield of Cu(OH)₂ = (174.6 / 292.5) × 100
Percentage yield of Cu(OH)₂ = 59.7%
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calculate the number of each atom in 2.5 gram of caco3
Answer:
[tex]2.5\; \rm g[/tex] of [tex]\rm CaCO_{3}[/tex] would contain:
Approximately [tex]1.5 \times 10^{22}[/tex] calcium atoms (approximately [tex]0.025\; \rm mol[/tex],)Approximately [tex]1.5 \times 10^{22}[/tex] carbon atoms (approximately [tex]0.025\; \rm mol[/tex],) andApproximately [tex]4.5 \times 10^{22}[/tex] oxygen atoms (approximately [tex]0.075\; \rm mol[/tex].)Explanation:
Look up the Avogadro constant: [tex]N_{\rm A} \approx 6.022 \times 10^{23}\; \rm mol^{-1}[/tex].
For example, "[tex]1\; \rm mol[/tex] of carbon atoms" would contain [tex]N_{\rm A}[/tex] carbon atoms (approximately [tex]6.022\times 10^{23}[/tex]) by definition.
Look up the relative atomic mass of carbon, calcium, and oxygen on a modern periodic table:
Calcium: [tex]40.078[/tex].Carbon: [tex]12.011[/tex].Oxygen: [tex]15.999[/tex].In other words, the mass of [tex]1\; \rm mol[/tex] of calcium atoms would be [tex]40.078\; \rm g[/tex]. The mass of [tex]1\; \rm mol\![/tex] of carbon atoms would be [tex]12.011\; \rm g[/tex], and the mass of [tex]1\; \rm mol \!\![/tex] of oxygen atoms would be [tex]15.999\; \rm g[/tex].
As the formula [tex]\rm CaCO_{3}[/tex] suggests, every formula unit of this ionic compound includes one calcium atom, one carbon atom, and three oxygen atoms. The formula mass of [tex]\rm CaCO_{3}\![/tex] would give the mass of every mole of [tex]\rm CaCO_{3}\!\![/tex] formula units.
Calculate the formula mass of [tex]\rm CaCO_{3}[/tex] from the relative atomic mass data:
[tex]\begin{aligned} & M({\rm CaCO_{3}}) \\ =\; & 40.078\; \rm g \cdot mol^{-1} \\ & + 12.011\; \rm g \cdot mol^{-1} \\ & + 3 \times (15.999\; \rm g \cdot mol^{-1}) \\ =\; & 100.086\; \rm g \cdot mol^{-1}\end{aligned}[/tex].
Calculate the number of [tex]\rm CaCO_{3}[/tex] formula units in that [tex]2.5\; \rm g[/tex] of this compound:
[tex]\begin{aligned}& n({\rm CaCO_{3}}) \\ =\; & \frac{m({\rm CaCO_{3}})}{M({\rm CaCO_{3}})} \\ =\; & \frac{2.5\; \rm g}{100.086\; \rm g \cdot mol^{-1}} \\ \approx\; & 0.025\; \rm mol\end{aligned}[/tex].
In other words, [tex]2.5\; \rm g[/tex] of [tex]\rm CaCO_{3}[/tex] would contain approximately [tex]0.025\; \rm mol[/tex] [tex]\rm CaCO_{3}\![/tex] formula units.
Again, there are one calcium atom, one carbon atom, and one oxygen atom in every [tex]\rm CaCO_{3}[/tex] formula unit. That approximately [tex]0.025\; \rm mol[/tex] [tex]\rm CaCO_{3}\![/tex] formula units would thus contain:
Approximately [tex]1 \times 0.025\; \rm mol = 0.025\; \rm mol[/tex] calcium atoms, Approximately [tex]1 \times 0.025\; \rm mol = 0.025\; \rm mol[/tex] carbon atoms, andApproximately [tex]3 \times 0.025\; \rm mol = 0.075\; \rm mol[/tex] oxygen atoms.Make use of the Avogadro constant to convert the numbers.
For example, the number of calcium atoms in that approximately [tex]0.025\; \rm mol[/tex] of calcium atoms would be:
[tex]\begin{aligned} & N({\text{calcium}) \\ = \; & n({\text{calcium}) \cdot N_{\rm A} \\ \approx \; & 0.025\; \rm mol \times 6.022\times 10^{23} \cdot mol^{-1} \\ \approx \; & 1.5 \times 10^{22} \end{aligned}[/tex].
Likewise, the number of carbon atoms in that approximately [tex]0.025\; \rm mol[/tex] of carbon atoms would be:
[tex]\begin{aligned} & N({\text{carbon}) \\ = \; & n({\text{carbon}) \cdot N_{\rm A} \\ \approx \; & 0.025\; \rm mol \times 6.022\times 10^{23} \cdot mol^{-1} \\ \approx \; & 1.5 \times 10^{22} \end{aligned}[/tex].
The number of oxygen atoms in that approximately [tex]0.075\; \rm mol[/tex] of oxygen atoms would be:
[tex]\begin{aligned} & N({\text{oxygen}) \\ = \; & n({\text{oxygen}) \cdot N_{\rm A} \\ \approx \; & 0.075\; \rm mol \times 6.022\times 10^{23} \cdot mol^{-1} \\ \approx \; & 4.5 \times 10^{22} \end{aligned}[/tex].
The density of gold is 19.3 g/cm3, and the density of iron pyrite is 5.02 g/cm3. What substance do you think you found?
Answer:
Step 1
1 of 2
V = m / D; V (nugget of gold) = 50 g / 5.0 g / cm 3 = 10 cm3; V (nugget of iron pyrite) = 50 g / 19 g / cm3 = 2.63 cm 3
Result
2 of 2
Due to these results you can conclude that nugget of gold which volume is 10 cm3 is bigger than volume of nugget of iron pyrite which is 2.63 cm3.
Explanation:
Al(s) + NaOH(aq) → Al(OH)3(s) + Na+(aq) displacement, or decomposition.
Answer:
Displacement
Explanation:
OH- in the reactant side of the reaction has switched from Na+ to Al 3+ in the product side. Decomposition is where a compound from the reactant side of the reaction would dissolve into elements in the product side, but that's not the case in this example.
What volume of 0.32M HCl is required to completely neutralize 50.0m mL of a 0.12M Mg(OH)2 solution?
Answer:
37.5 mL of HCl will be required for complete neutralization with magnesium hydroxide
Explanation:
[tex]{ \bf{2HCl _{(aq)} + Mg(OH)_{2(s)}→ MgCl_{2(s)} + 2H _{2} O_{(l)}}}[/tex]
we've to first get moles of magnesium hydroxide in 50.0 ml :
[tex]{ \sf{1 \: l \: of \: hydroxide \: contains \: 0.12 \: moles}} \\ { \sf{0.05 \: l \: of \: hydroxide \: contain \: (0.05 \times 0.12) \: moles}} \\ = { \underline{0.006 \: moles}}[/tex]
for complete neutralization:
[tex]{ \sf{1 \: moles \: of \: hydroxide\: react \: with \: 2 \: moles \: of \: acid}} \\ { \sf{0.006 \: moles \: react \: with \: \{0.006 \times 2 \}} \: moles} \\ = { \underline{0.012 \: moles \: of \: acid}}[/tex]
compare with acid molarity:
[tex]{ \sf{0.32 \: moles \: of \: acid \: occupy \: 1 \: litre}} \\ { \sf{0.012 \: moles \: of \: acid \: occupy \: ( \frac{0.012}{0.32}) \: litres }} \\ { \underline{ = 0.0375 \: litre \: \: = \: \: 37.5 \: ml}}[/tex]
Oceanography does not include the study of which of the following topics?
A.the relationship between the Sun, Moon, and Earth
B.the relationship between the atmosphere, Earth, and the oceans
C.the Earth's oceans
D.the plants and animals that live in the oceans
Fine the volume in dm³ and in Ml of a 0.505 mol/dm of NaCH required to react with 40Ml of 0.505 molar concentration of H2SO4
watch u know bout rolling down in the deep...
Which property is an intensive, rather than an extensive, property of matter?
A. volume
B. density
C. mass
D. length
Answer:
Density other dude is incorrect
Explanation:
Density is an intensive, rather than an extensive property of matter.
WHAT ARE INTENSIVE AND EXTENSIVE PROPERTIES?
The property of matter is divided into two categories namely: intensive and extensive properties.Intensive properties are those properties that are not dependent on the size or amount of the substance. Examples include density, temperature, boiling point, melting point etc. Extensive properties are those properties that depend on the size or amount of the substance involved. Examples are length, mass, volume weight etc.Therefore, density is an intensive, rather than an extensive property of matter.
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what was daltons motivation for proposing his model of the atom
Answer:
John Dalton based his partial pressures theory on the idea that only like atoms repel one another, Whereas unlike atoms appear to react indifferently. This notion was erroneous, but it helped to explain why each gas in a mixture behaved independently, serving the purpose of showing that atoms of all kinds are not alike.
If an inverse relationship exist between x and y, and k = 200, which of the following would be points on the graph?
-(200,0.1)
-(100, 0.2)
-(400, 0.5)
-(800, 0.4)
Answer:
The answer is (400, 0.5)
a hurricane that is rated category 5 would have which of the characteristics below?
A winds of 119-153 miles per an hour with little to no damage to structers
B winds of 154-177 miles per an hour with some damage to roofs doors or windows
C winds of 178-209 miles per hour with damage to homes
D winds of 251 miles per hour or higher with complete roof failure and possible complete structure failure
In rutherfords gold foil experiment, most alphas particles passed through the gold foil without deflection and were detected on the screen. What caused the particles to pass through without any deflection?
Answer:
The vast majority of the volume in an atom consists of the electron orbitals with electrons that have too little mass to deflect the alpha particles. So most pass right through the foil. Only the ones hitting the nucleus are deflected.
Explanation:
The whole reason Rutherford and his students conducted this experiment was to test the then current hypothesis that an atom consists of a homogeneous blend of atomic particles, known as the "plum pudding" model. If so, there should be no, or minimal, deflection of particles. That's not what they found. Since particles were found to be deflected, the group concluded that atoms must contain a large object, which became known as the nucleus.
Which compound is a carbohydrate. Nucleotide or protien or starch or amino acid
Which property altered during a chemical change is not altered during a physical change?
A. composition of the matter
B. temperature of the matter
C. volume of the matter
D. phase of the matter
The property altered during a chemical change that is not altered during a physical change is composition of the matter.
A physical change is one in which no new substance is formed and it is easily reversible.
A chemical change is one that is not easily reversible and no new substance is formed. It may be accompanied by absorption or evolution of heat.
The composition of a substance changes during a chemical change because bonds between atoms break and new bonds are formed. This does not occur during a physical change.
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Given the ion Al+3, what species would you expect to form with each of the following ions? OH- ___________ S2- ___________ PO43- _________
Ionic substances are formed by exchange of valencies between the ions that make up the substance.
The ion [tex]Al^3+[/tex] has a valency of +3. It exchanges this valency with other ions as it forms ionic substances.
In the case of OH- which has a valency of 1, the compound Al(OH)3 is formed.
In the case of [tex]S^2-[/tex] which has a valency of 2, the compound Al2S3 is formed.
In the case of [tex]PO4^3-[/tex] which has a valency of 3, the compound AlPO4 is formed.
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Name the reaction type & mechanism of the following reactions:
Answer:
MECHANISM:
1) The lone pair on oxygen attacks the H-Br molecule forming a hydronium ion.
2) Formation of carbocation.
3) Attack of Nucleophile Br − .
Explanation:
Answer:
DISPLACEMENT REACTION
COMBINATION REACTION
DOUBLE DISPLACEMENT REACTION
DECOMPOSITION REACTION
for the equation given, how many grams of methane will react with a 125g of oxygen
CH4 (g) + 2O2(g) -> CO2(g) +2H2O
Answer:
16
Explanation:
1) in the given equation CH₄+2O₂⇒ 2H₂O+CO₂;
M(CH₄)=12+4=16 (g/mol); M(O₂)=32 (g/mol);
2) if m(O₂)=125 gr., then ν(O₂)=m(O₂)/M(O₂)=125/32≈3.90625 (moles);
3) if in the given equation 'ν' of O₂ is 2, then the 'ν' of CH₄ is 1;
4) m(CH₄)=ν(CH₄)*M(CH₄)=1*16=16 (gr.).
Based on properties of elements in the periodic table, which element is the best conductor of thermal energy?
(1 point)
argon (Ar)
O germanium (Ge)
O vanadium (V)
polonium (Po)
Answer:
Vanadium (V) is the best conductor of thermal energy in this list
A 25.0 mL sample of 0.723 M HClO4 is titrated with a 0.273 M KOH solution. The H3O+ concentration after the addition of 66.2 mL of KOH is __________ M.
a. 0.723
b. 1.00 × 10-7
c. 2.81 × 10-13
d. 0.439
e. 0.273
What is the major organic product obtained from the reaction of ethylbenzene with nitronium ion, followed by benzylic oxidation?
a. 3-ethyl-aniline.
b. 4-nitro-benzoic acid.
c. 3-nitro-toluene.
d. 3-nitrobenzoic acid.
Answer:
b
Explanation:
Bonus 50 points if u got this answer correctly!!
When does a real gas behave like an ideal gas?
Answer:
[tex]at \: low \: pressure \: high \: temparature \\ thank \: you[/tex]
the electronic configuration of an atom of element X is 2,8,2 what will be the valency of the element x
The element is Magnesium
It has total 12 electronsit has electronic configuration
[tex]\\ \sf\longmapsto 2,8,2[/tex]
or
[tex]\\ \sf\longmapsto 1s^22s^22p^63s^2[/tex]
In valence cell it has 2 electrons and therefore its valency is 2
Which word equation best describes the following formula equation:
4Al + 302 + 2Al2O3?
Answer:
Explanation:
The equation is not correctly written, so let's start with that.
4Al + 3O2 ====> 2Al2O3?
Totally correct, including the balance numbers, would be
4moles aluminum + 3 moles Oxygen ===> 2 moles of Aluminum Oxide
Anything else is incomplete.
Answer:
aluminum reacts with oxygen to produce aluminum oxide.
I hope this helps
Axon diameters of cortico-cortical fibres in the human brain range from 0.16 um to 9 um in
length. Convert the length, 0.16 um, into meters using scientific notation. (Use the following
format: a x 10^b using the correct number of significant figures).
The conversion of units in science is achieved by the use of conversion factors.
A conversion factor is a number used to change a quantity from one unit to another.
The given unit here is the micrometer and we are expected to convert the quantity (mass in this case) from micrometer to meter.
The conversion factor to use is:
1 μm = [tex]1 * 10^-6 m[/tex]
Hence;
0.16 μm = 0.16 μm * 1 * 10^-6 m/ 1 μm
= 1.60 * 10^-7 m
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Which of the following statements best defines the actual yield of a reaction? (5 points)
Select one:
a.The amount of product measured after a reaction
b.The ratio of measured yield over theoretical yield
c.The maximum amount of product that can be obtained Incorrect
d.The ratio of measured yield over stoichiometric yield
Answer:
its d
. . . . . . . .. . . . . . . . .. .
A beggar wrapped himself with a few layers of newspaper on a cold winter night. This helped him to keep himself warm because:
(a) friction between the layers of newspaper produces heat.
(b) air trapped between the layers of newspaper is a bad conductor of heat.
(c) newspaper is a conductor of heat.
(d) newspaper is at a higher temperature than the temperature of the surrounding.
the one who gets its correct will be marked as the brainliest and also be followed by me!!
P.S. have a great Sunday :)
Answer:
(b)The air trapped between the layers of newspaper is a bad conductor of heat.Thus the layers of newspaper helped the beggar to keep himself warm
Hey there!
I believe your answer would be "Air trapped between the layers of newspaper is a bad conductor of heat."
Hope this helps!
Have a great day!
Whst type of molecule is
Dehydogenase
Answer:
A dehydrogenase is an enzyme .
Explanation:
Hope it helps
Mark me as Brainliest plz.
:DD
Are ALL acids and based
harmful to humans?
Do you think the video game "Geometry dash" is dead?
Answer:
HELL NO!
THAT'S MY CHILDHOOD GAME!
I LOVE THAT GAME!
IT SHOULD NOT BE DEAD!
IT SHOULD LIVE ON FOREVER!
#GemoteryDashShouldLiveOnForever!
[tex]AnimeVines[/tex]
What's the difference between science and science methods?
Answer:
Science is related only to sense-experience, and reason based on the former. It is an accumulation of systematic and verifiable knowledge. The scientific method is one and the same in all branches of knowledge. The unity of science, according to him, consists in its method, not in its material alone.
Explanation:
How much heat is absorbed by a 112.5 g sample of water when it is heated from 12.5 °C to 92.1 °C? (Specific heat capacity of water is 4.184 J/g °C)
Answer:
[tex]\boxed {\boxed {\sf37,467.72 \ Joules }}[/tex]
Explanation:
We are asked to find how much heat a sample of water absorbed. Since we are given the mass, temperature, and specific heat, we will use the following formula.
[tex]q=mc \Delta T[/tex]
The mass (m) of the sample is 112.5 grams. The specific heat capacity of water (c) is 4.184 Joules per gram degree Celsius. The difference in temperature (ΔT) is found by subtracting the initial temperature from the final temperature.
ΔT= final temperature - initial temperatureThe water was heated from 12.5 degrees Celsius to 92.1 degrees Celsius.
ΔT= 92.1 °C - 12.5 °C= 79.6°CNow we know three variables and can substitute them into the formula,
m= 112.5 g c= 4.184 J/g °C ΔT= 79.6 °C[tex]q= (112.5 \ g )(4.184 \ J/g \textdegree C)(79.6 \textdegree C)[/tex]
Multiply the first 2 numbers. Note the units of grams cancel.
[tex]q= (112.5 \ g *4.184 \ J/g \textdegree C)(79.6 \textdegree C)[/tex]
[tex]q= (112.5 *4.184 \ J/ \textdegree C)(79.6 \textdegree C)[/tex]
[tex]q= (470.7 \ J/ \textdegree C)(79.6 \textdegree C)[/tex]
Multiply again. This time the units of degrees Celsius cancel.
[tex]q= (470.7 \ J/ \textdegree C *79.6 \textdegree C)[/tex]
[tex]q= (470.7 \ J *79.6 )[/tex]
[tex]q= 37467.72 \ J[/tex]
37, 467.72 Joules of heat are absorbed by the sample fo water.