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Harman [31]
1 year ago
5

How many moles of carbon are in 7.87x10' carbon atoms?

Chemistry
1 answer:
larisa [96]1 year ago
6 0

Answer:

1.306 moles of C

Explanation:

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II. Pure magnesium metal is often found as ribbons and can easily burn in the presence of oxygen. When 3.86 g of magnesium ribbo
Leto [7]

Answer:

Excess=3.53g

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

2Mg(s)+O_2(g) \rightarrow 2MgO(s)

Next, we identify the limiting reactant by computing the moles of magnesium oxide yielded by 3.86 g of magnesium and 155 mL of oxygen at the given conditions via their 2:1:2 mole ratios and the ideal gas equation:

n_{MnO}^{by \ Mg}=3.86gMg*\frac{1molMg}{24.3gMg}*\frac{2molMgO}{2molMg}  =0.159molMgO\\\\n_{MnO}^{by \ O_2}=\frac{1atm*0.155L}{0.082\frac{atm*L}{molO_2*K}*275K} *\frac{2mol MgO}{1molO_2} =0.0137molMgO

It means that the limiting reactant is the oxygen as it yields the smallest amount of magnesium oxide. Next, we compute the mass of magnesium consumed the oxygen only:

m_{Mg}^{consumed}=0.0137molMgO*\frac{2molMg}{2molMgO} *\frac{24.3gMg}{1molMg} =0.334gMg

Thus, the mass in excess is:

Excess=3.86g-0.334g\\\\Excess=3.53g

Regards!

5 0
1 year ago
If an atom has sp3d2 hybridization in a molecule:
never [62]

Answer:

a. the maximum number of σ bonds that the atom can form is 4

b. the maximum number of p-p bonds that the atom can form is 2

Explanation:

Hybridization is the mixing of at least two nonequivalent orbitals, in this case, we have the mixing of one <em>s, 3 p </em> and <em> 2 d </em> orbitals. In hybridization the number of hybrid orbitals generated  is equal to the number of pure atomic orbital, so we have 6 hybrid orbital.

The shape of this hybrid orbital is octahedral (look the attached image) , it has 4 orbital located in the plane and 2 orbital perpendicular to it.

This shape allows the formation of maximum 4 σ bond, because σ bonds are formed by orbitals overlapping end to end.

And maximum 2 p-p bonds, because p-p bonds are formed by sideways overlapping orbitals. The atom can form one with each one of the orbitals located perpendicular to the plane.

4 0
1 year ago
Read 2 more answers
A solution was prepared by mixing 50.0 g
frez [133]

Answer:

4.78 %.

Explanation:

<em>mass percent is the ratio of the mass of the solute to the mass of the solution multiplied by 100.</em>

<em></em>

<em>mass % = (mass of solute/mass of solution) x 100.</em>

<em></em>

mass of MgSO₄ = 50.0 g,

mass of water = d.V = (0.997 g/mL)(1000.0 mL) = 997.0 g.

mass of the solution = mass of water + mass of MgSO₄ = 997.0 g + 50.0 g = 1047.0 g.

<em>∴ mass % = (mass of solute/mass of solution) x 100</em> = (50.0 g/1047.0 g) x 100 = <em>4.776 % ≅ 4.78 %.</em>

4 0
2 years ago
Exactly 56 grams of iron is mixed with 156 grams of oxygen. The elements are heated and they react. What best describes which re
Mars2501 [29]

Answer:

Explanation:

The chemical expression for the reaction between iron and oxygen is:

4Fe(s)  +  3O₂ (g)  \to 2Fe₂O₃ (s)

The number of moles of Fe = mass of Fe/ molecular mass of Fe

The number of moles of Fe = 56 g/ 55.845 g/mol

The number of moles of Fe = 1.002 moles of Fe

The number of moles of oxygen = mass of oxygen/ molecular mass of oxygen

The number of moles of oxygen = 156 g /32 g/mol

The number of moles of oxygen = 4.875 moles of oxygen

Assume that Fe is the limiting reactant, the number of Fe₂O₃ can be calculated as:

moles of Fe₂O₃ = 1.002 mole of Fe × 2 moles of Fe₂O₃/ 4 moles of Fe

moles of Fe₂O₃ = 0.501 mole of Fe₂O₃

Assume that O₂ is the limiting factor, the number of Fe₂O₃ is:

moles of Fe₂O₃ = 4.875 moles of O₂ × 2 moles of Fe₂O₃/ 3 moles of O₂

moles of Fe₂O₃ = 3.25 mole of Fe₂O₃

Thus, after the reaction is complete, Fe and O₂ contain different moles of Fe₂O₃. Only Fe gets consumed in the reaction and it is the limiting factor.

8 0
1 year ago
Small quantities of h2 gas can be collected by adding hcl to zn. a sample of 195 ml of h2 gas was collected over water at 25 c a
Umnica [9.8K]
<span>15.4 milligrams The ideal gas law is PV = nRT where P = pressure of the gas V = volume of the gas n = number of moles of gas R = Ideal gas constant (8.3144598 L*kPa/(K*mol) ) T = absolute temperature. So let's determine how many moles of gas has been collected. Converting temperature from C to K 273.15 + 25 = 298.15 K Converting pressure from mmHg to kPa 753 mmHg * 0.133322387415 kPa/mmHg = 100.3917577 kPa Taking idea gas equation and solving for n PV = nRT PV/RT = n n = PV/RT Substituting known values n = PV/RT n = (100.3917577 kPa 0.195 L) / (8.3144598 L*kPa/(K*mol) 298.15 K) n = (19.57639275 L*kPa) / (2478.956189 L*kPa/(mol) ) n = 0.007897031 mol So we have a total of 0.007897031 moles of gas particles. Now let's get rid of that percentage that's water vapor. The percentage of water vapor is the vapor pressure of water divided by the total pressure. So 24/753 = 0.03187251 The portion of hydrogen is 1 minus the portion of water vapor. So 1 - 0.03187251 = 0.96812749 So the number of moles of hydrogen is 0.96812749 * 0.007897031 mol = 0.007645332 mol Now just multiple the number of moles by the molar mass of hydrogen gas. Start with the atomic weight. Atomic weight hydrogen = 1.00794 Molar mass H2 = 1.00794 * 2 = 2.01588 g/mol Mass H2 = 2.01588 g/mol * 0.007645332 mol = 0.015412073 g Rounding to 3 significant figures gives 0.0154 g = 15.4 mg</span>
7 0
2 years ago
Read 2 more answers
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