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3 years ago

HURRY PLEASE

Chemistry

1 answer:

Digiron [165]3 years ago

6 0

Answer:

The Combined Gas Law shows that the pressure of a gas is inversely proportional to volume and directly proportional to temperature. Avogadro's Law shows that volume or pressure is directly proportional to the number of moles of gas. Putting these together leaves us with the following equation:

P1×V1T1×n1=P2×V2T2×n2(11.9.1)

As with the other gas laws, we can also say that (P×V)(T×n) is equal to a constant. The constant can be evaluated provided that the gas being described is considered to be ideal.

The Ideal Gas Law is a single equation which relates the pressure, volume, temperature, and number of moles of an ideal gas. If we substitute in the variable R for the constant, the equation becomes:

P×VT×n=R(11.9.2)

The Ideal Gas Law is conveniently rearranged to look this way, with the multiplication signs omitted:

PV=nRT(11.9.3)

The variable R in the equation is called the ideal gas constant.

Explanation:

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Answer the following for the reaction: 3AgNO3(aq)+Na3PO4(aq)→Ag3PO4(s)+3NaNO3(aq)

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Answer:1) Volume of $AgNO_3$ required is 55.98 mL.

2) 0.62577 grams of $Ag_3PO_4$ is produced.

Explanation:

$3AgNO_3(aq)+Na_3PO_4(aq)\rightarrow Ag_3PO_4(s)+3NaNO_3(aq)$

1) Molarity of $AgNO_3,M_1=0.225 M$

Volume of $AgNO_3.V_1=?$

Molarity of $Na_3PO_4,M_2=0.135 M$

Volume of $Na_3PO_4,V_2=31.1 mL=0.0311 L$

$Molarity=\frac{\text{number of moles}}{\text{volume of solution in liters}}$

$\text{number of moles }Na_3PO_4=M_2\times V_2=0.135 mol/L\times 0.0311 L=0.0041985 moles$

According to reaction, 1 mole of $Na_3PO_4$ reacts with 3 mole of $AgNO_3$ , then, 0.0041985 moles of $Na_3PO_4$ will react with:

$\frac{3}{1}\times 0.0041985$ moles of $AgNO_3$ that is 0.0125955 moles.

$M_1=0.225 M=\frac{\text{number of moles of }AgNO_3}{V_1}$

$V_1=\frac{0.0125955 moles}{0.225 M}=0.05598 L=55.98 mL$

Volume of $AgNO_3$ required is 55.98 mL.

$Molarity=0.195 M=\frac{\text{number of moles}}{\text{volume of solution in liters}}$

Number of moles of $AgNO_3=0.195\times 0.023 L=0.004485 moles$

According to reaction, 3 moles of $AgNO_3$ gives 1 mole of $Ag_3PO_4$ , then 0.004485 moles of $AgNO_3$ will give: $\frac{1}{3}\times 0.004485$ moles of $Ag_3PO_4$ that is 0.001495 moles.

Mass of $Ag_3PO_4$ =

Moles of $Ag_3PO_4$ × Molar Mass of $Ag_3PO_4$

= 0.001495 moles × 418.58 g/mol = 0.62577 g

0.62577 grams of $Ag_3PO_4$ is produced.

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