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kolezko [41]
3 years ago
11

How many moles of Co2 will be produced if 9.0 moles of O2 completely react?

Chemistry
1 answer:
ale4655 [162]3 years ago
8 0
 <span>It depends on what was being combusted. 

Supposing the simplest case, where carbon is being burnt: 

C + O2 → CO2 

(9.0 mol O2) x (1 mol CO2 / 1 mol O2) = 9.0 mol CO2 

The answer would be different if the fuel contained any oxygen atoms, for example, alcohol.</span><span>
</span>
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Dissolving 5.28 g of an impure sample of calcium carbonate in hydrochloric acid produced 1.14 L of carbon dioxide at 20.0 °C and
swat32

Answer:

\%\ mass\ of\ CaCO_3=93.37\ \%

Explanation:

Given that:

Pressure = 791 mmHg

Temperature = 20.0°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (20 + 273.15) K = 293.15 K  

T = 293.15 K  

Volume = 100 L

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 62.3637 L.mmHg/K.mol  

Applying the equation as:

791 mmHg × 1.14 L = n × 62.3637 L.mmHg/K.mol  × 293.15 K  

⇒n of CO_2 produced =  0.0493 moles

According to the reaction:-

CaCO_3 + 2 HCl\rightarrow CaCl_2 + H_2O + CO_2

1 mole of carbon dioxide is produced 1 mole of calcium carbonate reacts

0.0493 mole of carbon dioxide is produced 0.0493 mole of calcium carbonate reacts

Moles of calcium carbonate reacted = 0.0493 moles

Molar mass of CaCO_3 = 100.0869 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

0.0493\ mol= \frac{Mass}{100.0869\ g/mol}

Mass_{CaCO_3}=4.93\ g

Impure sample mass = 5.28 g

Percent mass is percentage by the mass of the compound present in the sample.

\%\ mass\ of\ CaCO_3=\frac{Mass_{CaCO_3}}{Total\ mass}\times 100

\%\ mass\ of\ CaCO_3=\frac{4.93}{5.28}\times 100

\%\ mass\ of\ CaCO_3=93.37\ \%

3 0
3 years ago
Cci(4) will be soluble in
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LIKE DISSOLVES LIKE. Since Ccl4 is non-polar, it'll be soluble in any non-polar solvent. Hope this helps you!
8 0
3 years ago
An element has ccp packing with a face-centered cubic unit cell. its density is 8920 kg/m3 and the unit cell volume is 4.72 x 10
ziro4ka [17]
<span>63.4 g/mol
   First, let's determine how many atoms per unit cell in face-centered cubic. There is 8 corners, each of which has 1 atom, and each of those atoms is shared between 8 other unit cells. So 8*1/8 = 1 atom per unit cell. Additionally, there are 6 faces, each of which has 1 atom that's shared between 2 unit cells. So 6*1/2 = 3 atoms per unit cell. So each unit cell has the mass of 1+3 = 4 atoms. Since there is 1000 liters per cubic meter, the mass per liter is 8920 kg/1000 = 8.920 kg/L. Now the mass per unit cell is 8920 g * 4.72x10^-26 = 4.21024x10^-22 g per unit cell. The mass per atom is 4.21024x10^-22 g / 4 = 1.05256x10^-22 g/atom, Finally, multiply by Avogadro's number, getting 1.05256x10^-22 g/atom * 6.0221409x10^23 atom/mol = 63.38664625704 g/mol.
   Rounding to 3 significant digits gives 63.4 g/mol.</span>
5 0
3 years ago
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