Which pair of reactants for a Grignard reaction does not give 2-phenylbutan-2-ol after an aqueous workup?

Explanation:

For Mg, (1.2 g Mg/24 g Mg) = 0.05 mol Mg.

For Cl, (3.55 g Cl/35.5 g Cl) = 0.1 mol Cl

So the ratio now is

Mg:Cl = 0.05 : 0.1 = 1:2

I got the 1:2 ratio by dividing both by the smallest number, which is 0.05 mol. Therefore, the empirical for formula of the substance is [tex]MgCl_2[/tex]

The empirical formula of the compound formed when magnesium reacts with chlorine is MgCl₂.

To determine the empirical formula of the compound formed when magnesium (Mg) reacts with chlorine (Cl), we need to find the ratio of the elements in terms of moles.

First, we calculate the number of moles of each element:

Moles of Mg = Mass of Mg / Molar mass of Mg

Moles of Mg = 1.2 g / 24 g/mol

Moles of Mg = 0.05 mol

Moles of Cl = Mass of Cl / Molar mass of Cl

Moles of Cl = 3.55 g / 35.5 g/mol

Moles of Cl = 0.1 mol

Next, we determine the simplest whole-number ratio between the moles of Mg and Cl:

Mg:Cl = 0.05 mol : 0.1 mol

To simplify the ratio, we divide both values by the smaller value (0.05 mol in this case):

Mg:Cl = 0.05 mol / 0.05 mol : 0.1 mol / 0.05 mol

Mg:Cl = 1 : 2

Therefore, the empirical formula of the compound formed when magnesium reacts with chlorine is MgCl₂.

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The groups are alkali metals (I), alkaline earth metal (II) , Halogen (VII), and noble gases (VIII).

The representative elements, also known as the primary group elements, are found in Group A. As a result, the columns beginning with A (IA through VIIIA) are referred to as the primary group elements. The periodic table's columns were referred to as groups.

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Answer:

787.401575 inches

Answer:

5) C

6) D

7) D

8) C

Explanation:

Answer:

पेरिोदिछ ताबले माछकिनौय

Answer:

on the right side of the periodic table

Answer:

2NaOH + CO2 -> Na2CO3 + H2O

1) Find the moles of each substance

[tex]\eq n(NaOH)=\frac{35.0}{22.99+16.00+1.008\\}\ =\frac{35.0}{39.998} \ = 0.8750437522 moles\\n(CO_{2} ) = \frac{17.5}{12.01+32.00} = \frac{17.5}{44.01} = 0.3976369007 moles\\[/tex]

2) Determine the limitting reagent

[tex]\\NaOH = \frac{0.8750437522}{2} = 0.4375218761\\\\[/tex]

∴ Carbon dioxide is limitting as it has a smaller value.

3) multiply the limiting reagent by the mole ratio of unknown over known

n(H2O ) = 0.3976369007 × 1/2

             = 0.1988184504 moles

4) Multiply the number of moles by the molar mass of the substance.

m = 0.1988184504 × (1.008 × 2 + 16.00)

   = 0.1988184504 × 18.016

   = 3.581913202 g

Explanation:

The correct answer is = 3.581913202 g.

n(H2O ) = 0.3976369007 × 1/2

             = 0.1988184504 moles

  m = 0.1988184504 × (1.008 × 2 + 16.00)

      = 0.1988184504 × 18.016

      = 3.581913202 g.

H2O is the molecular formula of water, also called Dihydrogen monoxide. H2O symbolises two hydrogen atoms and one oxygen atom.

Water (H2O) is made from 2 atoms of hydrogen and 1 atom of oxygen. A mole of water molecules would be 2 moles of hydrogen atoms plus 1 mole of oxygen atoms. From the periodic table, we see the atomic weight of hydrogen is 1.0079 and the atomic weight of oxygen is 15.9994.

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Answer:

[tex]\boxed {\boxed {\sf A. \ 12.5 \ grams}}[/tex]

Explanation:

We are asked to find the mass of 2.34 × 10²³ atoms of sulfur (S)

First, we convert atoms to moles using Avogadro's Number of 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. For this question, the particles are atoms of sulfur. There are 6.022 × 10²³ atoms of sulfur in 1 mole of sulfur.

We will convert using dimensional analysis, so we must set up a ratio.

[tex]\frac {6.022 \times 10^{23} \ atoms \ S}{1 \ mol \ S}[/tex]

We are converting 2.34 × 10²³ atoms of sulfur to moles, so we multiply the ratio by this value.

[tex]2.34 \times 10^{23} \ atoms \ S* \frac {6.022 \times 10^{23} \ atoms \ S}{1 \ mol \ S}[/tex]

Flip the ratio so the units of atoms of sulfur cancel.

[tex]2.34 \times 10^{23} \ atoms \ S* \frac{1 \ mol \ S} {6.022 \times 10^{23} \ atoms \ S}[/tex]

[tex]2.34 \times 10^{23}* \frac{1 \ mol \ S} {6.022 \times 10^{23}}[/tex]

[tex]\frac{2.34 \times 10^{23}} {6.022 \times 10^{23}} \ mol \ S[/tex]

[tex]0.3885752242 \ mol \ S[/tex]

Next, we convert moles to grams using the molar mass. This is the mass of 1 mole of a substance. The value is found on the Periodic Table as the atomic mass, but the units are grams per mole instead of atomic mass units. Look up sulfur's molar mass.

Set up another ratio.

[tex]\frac {32.07 \ g \ S}{ 1\ mol \ S}[/tex]

Multiply by the number of moles we calculated.

[tex]0.3885752242 \ mol \ S *\frac {32.07 \ g \ S}{ 1\ mol \ S}[/tex]

The units of moles of sulfur cancel.

[tex]0.3885752242 *\frac {32.07 \ g \ S}{ 1}[/tex]

[tex]0.3885752242 * {32.07 \ g \ S}[/tex]

[tex]12.46160744 \ g \ S[/tex]

The original value of atoms (2.34 × 10²³) has 3 significant figures (2,3, 4), so our answer must have the same. For the number we calculated, that is the tenths place. The 6 in the hundredth place tells us to round the 4 up to a 5.

[tex]12.5 \ g \ S[/tex]

There are 12.5 grams of sulfur in 2.34 × 10²³ atoms of sulfur.

Answer:

2.98 g CO2

Explanation:

1. Balance the chemical equation

1 Fe2O3 + 3CO ->  2Fe + 3CO2

2. Use periodic table to find mass from each CO2 and Fe2O3  

3. Calculate

3.60 g Fe2O3 x [tex]\frac{1 mol Fe2O3}{159.69 g Fe2O3}x\frac{3 mol CO2}{1 mol Fe2O3}[/tex] x[tex]\frac{44.01 g CO2}{1 mol CO2}[/tex] =  2.98 g CO2

The grams of CO, produced when 3.60 g of Fe 0 reacts with excess CO is 2.98 grams Co2.

Mass is the quantity of matter of a physical body.

The balanced equation is

[tex]\rm Fe_2O_3 + CO - > Fe + CO_2[/tex]

The calculation is

[tex]3.60 g \;Fe_2O_3 \times \dfrac{ 1 mol\;Fe_2O_3}{159.69} \times \dfrac{3 mol CO_2}{1 mol \;Fe_2O_3 } \times \dfrac{44.01\;CO_2}{1 mol\; CO_2} = 2.98 g\;CO_2[/tex]

Thus, the grams of CO, produced when 3.60 g of Fe 0 reacts with excess CO is 2.98 grams Co2.

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Explanation:

Ammonia is produced by the reaction of nitrogen and hydrogen according to the equation

N2(g) + 3H2(g) → 2NH3(g)

Calculate the mass of ammonia produced when 27.0 g of nitrogen react with 12.0 g of hydrogen.

g NH3

Which is the excess reactant and how much of it will be left over when the reaction is complete?

hi <3

the cloudiness is caused by the production of carbon dioxide

hope this helps :)

Answer:

Lime water is calcium hydroxide with formula

[tex]{ \sf{Ca(OH)_2}}[/tex]

when calcium hydroxide reacts with the gas [ carbon dioxide ]

[tex]{ \sf{Ca(OH)_{2(aq)} + CO_{2(g)} → CaCO_{3(s)} + H_2O_{(l)}}}[/tex]

Calcium carbonate is white cloudy precipitate

Answer:

The atomic number of an atom [ number of protons ] is the number of electrons of an atom.

[tex].[/tex]

Answer:

fddoiiijjjjjjjhhggggrrddffgghuuiii

Answer:

50000 g

Explanation:

1 kilogram (kg) is equal to 1000 grams (g).

1 kg = 1000 g

The mass m in grams (g) is equal to the mass m in kilograms (kg) times 1000:

m(g) = m(kg) × 1000

Answer:

because it has the same number of protons as electrons.

Answer:

The speed of a roller coaster increases as gravity pulls it downhill and is greatest at its lowest point. Viewed in terms of energy, the roller-coaster-Earth system's gravitational potential energy is converted to kinetic energy

Explanation:

a ball held in the air is the example of a change from gravitational potential energy to kinetic energy.

The final product of the reaction is 4-amino-3-hydroxy-1-acetylbenzene, also known as N-acetylanthranilic acid.

A catalyst is a substance that accelerates or reduces the temperature or pressure required to initiate a chemical reaction without being consumed in the process.

The Friedel-Crafts acylation reaction occurs when benzene reacts with acetic anhydride in the presence of an acid catalyst.

The acid catalyst activates the acetic anhydride to form an acylium ion intermediate in this reaction. This intermediate reacts with the benzene ring, giving it an acetyl group.

On carbon 1, there is a hydroxy group, and on carbon 4, there is an NH2 group.

Both of these functional groups are activating groups, which means they donate electrons to the benzene ring and increase its reactivity to electrophilic aromatic substitution reactions like Friedel-Crafts acylation.

Acetic anhydride adds an acetyl group to the benzene ring in a regioselective manner, which means it only adds to one position on the ring.

Thus, in this case, it adds to carbon 1, which is more reactive than carbon 4 due to the presence of the hydroxy group.

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Answer:

Explanation:

0.0

Lanthanoids and actinoids are transition elements of the periodic table. These elements are lustrous and malleable and can conduct thermal energy and electricity. Thus, option A is correct.

Lanthanoids and actinoids are the terms used to refer to the elements of the f-block of the periodic table that are placed separately due to the difference in their electronic structures.

These elements show the property somewhat similar to that of the metals and hence are lustrous and malleable. Due to the transition electron phase, they are used in optical devices and cardiac pacemakers.

The lanthanoids and actinoids are good conductors of heat and electricity as their electrons can be transferred and carry charges like metals.

Therefore, option A. the lanthanoids and actinoids are conductors and are malleable.

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Answer:

Hi, There! the Correct Answer is below!

C.)interactions between humans and living and non-living things in the environment

Explanation:

Environmental scientists study subjects like the understanding of earth processes, evaluating alternative energy systems, pollution control and mitigation, natural resource management, and the effects of global climate change. ... Environmental scientists bring a systems approach to the analysis of environmental problems.

____________________________________________________________

Have a wonderful Day :)

~Darkspirit~

Answer:

interactions between human and living and non living things in the environment

Thermometers are used to measure temperatures. The temperature of the hot junction of the thermocouple is [tex]520^oC[/tex]

Given that:

[tex]EMF = 15mv[/tex]

[tex]T_1 =20^oC[/tex] --- Temperature at cold junction

[tex]s = 0.03mv/^oC[/tex] ---- Sensitivity

The temperature [tex]T_2[/tex] at the hot junction is calculated as follows:

[tex]T_2 = \frac{EMF}{s} + T_1[/tex]

So, we have:

[tex]T_2 = \frac{15mv}{0.03mv/^oC} + 20^oC[/tex]

[tex]T_2 = \frac{15}{0.03/^oC} + 20^oC[/tex]

[tex]T_2 = \frac{15}{0.03}^oC + 20^oC[/tex]

[tex]T_2 = 500^oC + 20^oC[/tex]

[tex]T_2 = 520^oC[/tex]

Hence, the temperature of the hot junction of the thermocouple is [tex]520^oC[/tex]

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Answer:

ज

Explanation:

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Answer:

In this type of solution, perchloric acid is a strong acid which means that its conjugate base is extremely weak. As such, the solution of HClO4 H C l O 4 and NaClO4 N a C l O 4 is not a buffer solution.

Explanation:

We can measure the weight or mass of something by using beam balance.

oxide in a chemical reaction

CuFe2O4 are the products of combining iron (II) and Copper (II) oxide in a Chemical reaction.

The conversion of one or more chemicals into new one or more chemicals is called chemical reaction.

Example: iron reacts with oxygen to form rust.

The final substance of chemical reaction is called product.

Example: CuFe2O4 is the final substance called product when iron (II) and Copper (II) oxide react with each other.

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The Earth’s tilt changes its position relative to the stars and constellations as the Earth rotates around its axis.

Answer:

Not sure but I think it is C!

Explanation:

Answer:

[tex]\boxed {\boxed {\sf 2.001 \ mol \ Al_2O_3}}[/tex]

Explanation:

We are asked to convert grams of aluminum oxide to moles. We will do this using the molar mass or mass of 1 mole of a substance. The values are found on the Periodic Table. They are equal to the atomic masses, but the units are grams per mole instead of atomic mass units.

We have the compound aluminum oxide or Al₂O₃. We must look up the molar masses of the individual elements: aluminum and oxygen.

The formula of the compound has subscripts (a subscript of 2 after Al and subscript of 3 after O). There are 2 moles of aluminum and 3 moles of oxygen in 1 mole of aluminum oxide. We must multiply the molar mass by its subscript.

We convert using dimensional analysis, so we set up a ratio using the molar mass.

[tex]\frac {101.960076 \ g \ Al_2O_3}{ 1 \ mol \ Al_2O_3}[/tex]

We are converting 204.0 grams to moles, so we multiply by this value.

[tex]204.0 \ g \ Al_2O_3 *\frac {101.960076 \ g \ Al_2O_3}{ 1 \ mol \ Al_2O_3}[/tex]

Flip the ratio so the units of grams of aluminum oxide cancel.

[tex]204.0 \ g \ Al_2O_3 *\frac { 1 \ mol \ Al_2O_3}{101.960076 \ g \ Al_2O_3}[/tex]

[tex]204.0 *\frac { 1 \ mol \ Al_2O_3}{101.960076}[/tex]

[tex]\frac {204.0}{101.960076} \ mol \ Al_2O_3[/tex]

[tex]2.00078313 \ mol \ Al_2O_3[/tex]

The original measurement of grams has 4 significant figures, so our answer must have the same. For the number we found, that is the thousandth place. The 7 in ten-thousandth place tells us to round the 0 up to a 1.

[tex]2.001 \ mol \ Al_2O_3[/tex]

204.04 grams of aluminum oxide is equal to approximately 2.01 moles of aluminum oxide.