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3 years ago

Write a balanced equation for the following reaction:

Chemistry

1 answer:

tiny-mole [99]3 years ago

6 0

Answer:

3AgCl + Na₃PO₄ —> 3NaCl + Ag₃PO₄

The coefficients are 3, 1, 3, 1

Explanation:

From the question given above, the following data were:

Silver chloride reacts with sodium phosphate to yield sodium chloride and silver phosphate. This can be written as follow:

AgCl + Na₃PO₄ —> NaCl + Ag₃PO₄

The above equation can be balanced as follow:

AgCl + Na₃PO₄ —> NaCl + Ag₃PO₄

There are 3 atoms of Na on the left side and 1 atom on the right side. It can be balance by putting 3 in front of NaCl as shown below:

AgCl + Na₃PO₄ —> 3NaCl + Ag₃PO₄

There are 3 atoms of Cl on the right side and 1 atom on the left. It can be balance by putting 3 in front of AgCl as shown below:

3AgCl + Na₃PO₄ —> 3NaCl + Ag₃PO₄

Thus, the equation is balanced.

The coefficients are 3, 1, 3, 1

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2 years ago

A balloon contains 5 L of oxygen at a pressure of 90 kPa. How much oxygen will the balloon contain if the pressure is lowered to

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3 years ago

Does changing the number of neutrons change the identity of the element you have built ?

worty [1.4K]

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Explanation:

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3 years ago

If you pump air into cycle tyre a slight warming effect is noticed at valve stem why

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3 years ago

mixture of N 2 And H2 Gases weighs 13.22 g and occupies a volume of 24.62 L at 300 K and 1.00 atm.Calculate the mass percent of

anygoal [31]

Answer: The mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

Explanation:

To calculate the number of moles, we use the equation given by ideal gas equation:

PV = nRT

where,

P = Pressure of the gaseous mixture = 1.00 atm

V = Volume of the gaseous mixture = 24.62 L

n = number of moles of the gaseous mixture = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = Temperature of the gaseous mixture = 300 K

Putting values in above equation, we get:

$1.00atm\times 24.62L=n_{mix}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 300K\\\\n_{mix}=\frac{1.00\times 24.62}{0.0821\times 300}=0.9996mol$

We are given:

Total mass of the mixture = 13.22 grams

Let the mass of nitrogen gas be 'x' grams and that of hydrogen gas be '(13.22 - x)' grams

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

For nitrogen gas:

Molar mass of nitrogen gas = 28 g/mol

$\text{Moles of nitrogen gas}=\frac{x}{28}mol$

For hydrogen gas:

Molar mass of hydrogen gas = 2 g/mol

$\text{Moles of hydrogen gas}=\frac{(13.22-x)}{2}mol$

Equating the moles of the individual gases to the moles of mixture:

$0.9996=\frac{x}{28}+\frac{(13.22-x)}{2}\\\\x=12.084g$

To calculate the mass percentage of substance in mixture we use the equation:

$\text{Mass percent of substance}=\frac{\text{Mass of substance}}{\text{Mass of mixture}}\times 100$

Mass of the mixture = 13.22 g

For nitrogen gas:

Mass of nitrogen gas = x = 12.084 g

Putting values in above equation, we get:

$\text{Mass percent of nitrogen gas}=\frac{12.084g}{13.22g}\times 100=91.41\%$

For hydrogen gas:

Mass of hydrogen gas = (13.22 - x) = (13.22 - 12.084) g = 1.136 g

Putting values in above equation, we get:

$\text{Mass percent of hydrogen gas}=\frac{1.136g}{13.22g}\times 100=8.59\%$

Hence, the mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

5 0

3 years ago