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3 years ago

Aluminum reacts with chlorine gas to form aluminum chloride

Chemistry

1 answer:

VikaD [51]3 years ago

8 0

Answer: 48.95g

Explanation:

no. of moles of Cl2 = 39/(2*35.5) = 0.55 mol

no. of moles of Al = 34/27 = 1.26 mol

hence, aluminium is in excess so we'll do calculation using no. of moles of Cl2 as it will be the only reactant to be used up completely. So,

no of moles of AlCl3 = 2/3 * (0.55) = 0.367 mol

hence amount of AlCl3 = 0.367 * (27+3*35.5) = 48.95g

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3 years ago

The density of an unknown gas is 4. 20 grams per liter at 3. 00 atmospheres pressure and 127 °c. what is the molecular weight of

Kryger [21]

The molecular weight of this gas will be 45 g/mol .

The state equilibrium equation for a fictitious perfect gas is known as the ideal gas law, sometimes known as the generic gas equation. Although it has significant drawbacks, it represents a decent approximation of the activity of many gases under various conditions.

Ideal gas law can be expressed as:

PV =nRT

Calculation of molecular weight by using ideal as law.

Given data:

P = 3 atm

T = 127 °c

Density = 4. 20 grams per liter

PV =nRT

where p is pressure , T is temperature and R is gas constant.

PV = gram / molecular weight RT

Molecular weight = (g/v)( 1/P) RT

Putting the given data in above equation.

Molecular weight =4.20 × 1/ 3× 400 × 0.0831

Molecular weight = 45 g/mol.

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When water freezes it forms a lattice pattern and

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3 years ago

How many moles of CO2 are emitted into the atmosphere when 25.6 g C8H18 is burned

german

When 25.6 g $C_{8} H_{18}$ is burned, the moles of $CO_2$ emitted into the atmosphere is 1.80 mol

<u>Explanation:</u>

The molar mass of octane $C_{8} H_{18}$ is calculated as 8(12) +18(1) = 114 g/mol.

Where atomic mass of Carbon is C=12, and Hydrogen H=1.The sample contains 25.6 g of octane, Therefore

$\frac{25.6 g}{114 g / m o l}=0.225 \mathrm{mol}$

The combustion of octane can be written as:

$2 C_{8} H_{18}+25 O_{2} \rightarrow 16 C O_{2}+18 H O_{2}$

The mole ratio can be represented as:

$C O_{2}: C_{8} H_{18}$ = 16:2 = 8:1.

Hence, the quantity of carbon dioxide emitted is:

$\frac{8}{1} \times 0.225 \mathrm{mol}$ = 1.80 mol

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Answer:

Electromagnetic spectrum

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