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Lilit [14]
3 years ago
10

2. The temperature of a gas is increased from 125⁰C to 182⁰C inside of a rigid container. The original pressure of the gas was 1

.22atm, what will the pressure of the gas be after the temperature change?
Chemistry
2 answers:
Lina20 [59]3 years ago
7 0

P1/T1 = P2/T2

125⁰C = 398.15 k

182⁰C = 455.15 k

1.22/398.15 = p2/455.15


p2= 1.39atm

the pressure of the gas be after the temperature change is 1.39 atm

jekas [21]3 years ago
5 0

Answer:

1.4 atm

Explanation:

According to Gay Lussac’s law, the pressure of gas is inversely proportional t its temperature, when the volume is constant

Initial temperature = 125 + 273 = 398 K

Final temperature = 182 + 273 = 455 K

Initial pressure = 1.22 atm

\frac{P1}{T1} = \frac{P2}{T2}

1.22 atm / 398 K = P₂/ 455 K

P₂ = \frac{1.22 atm x 455 K}{398 K} = 1.4 atm

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How many liters of hydrogen gas will be produced at STP from the reaction of 7.179×10^23 atoms of magnesium with 54.219g of phos
Alexeev081 [22]

Answer: The volume of hydrogen gas produced will be, 12.4 L

Explanation : Given,

Mass of H_3PO_4 = 54.219 g

Number of atoms of Mg = 7.179\times 10^{23}

Molar mass of H_3PO_4 = 98 g/mol

First we have to calculate the moles of H_3PO_4 and Mg.

\text{Moles of }H_3PO_4=\frac{\text{Given mass }H_3PO_4}{\text{Molar mass }H_3PO_4}

\text{Moles of }H_3PO_4=\frac{54.219g}{98g/mol}=0.553mol

and,

\text{Moles of }Mg=\frac{7.179\times 10^{23}}{6.022\times 10^{23}}=1.19mol

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

3Mg+2H_3PO_4\rightarrow Mg(PO_4)_2+3H_2

From the balanced reaction we conclude that

As, 3 mole of Mg react with 2 mole of H_3PO_4

So, 0.553 moles of Mg react with \frac{2}{3}\times 0.553=0.369 moles of H_3PO_4

From this we conclude that, H_3PO_4 is an excess reagent because the given moles are greater than the required moles and Mg is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of H_2

From the reaction, we conclude that

As, 3 mole of Mg react to give 3 mole of H_2

So, 0.553 mole of Mg react to give 0.553 mole of H_2

Now we have to calculate the volume of H_2  gas at STP.

As we know that, 1 mole of substance occupies 22.4 L volume of gas.

As, 1 mole of hydrogen gas occupies 22.4 L volume of hydrogen gas

So, 0.553 mole of hydrogen gas occupies 0.553\times 22.4=12.4L volume of hydrogen gas

Therefore, the volume of hydrogen gas produced will be, 12.4 L

4 0
3 years ago
The formula weight of magnesium hydroxide is________ amu.<br> amu.
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2 years ago
How many grams of ammonia must you start with to make 900.00 l of a 0.140 m aqueous solution of nitric acid? assume all the reac
bogdanovich [222]
You need the set of reactions that goes from ammonia to nitric acid.
<span>
1) 4NH3(g)+5O2(g)-->4NO(g)+6H2O(g)

2) 2NO(g)+O2(g)-->2NO2(g)

3) 3NO2(g)+H2O(l)-->2HNO3(aq)+NO(g)

State the ratio of moles of HNO3 to NH3:

4 moles of NH3 produce 4 mole of NO,

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4 moles of NO2 produce 4 * (2 / 3) moles of HNO3 = 8/3 moles of HNO3.

=> (8/3) moles HNO3 : 4 moles NH3

Calculate the number of moles of HNO3 in 900.00 l of 0.140 M solution

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Use proportions:

(</span><span>8/3) moles HNO3 / 4 moles NH3 = 126 moles HNO3 / x

=> x = 126 moles HNO3 * 4 moles NH3 / (8/3 moles HNO3) = 189 moles NH3

Convert moles to grams:

molar mass NH3 = 14 g/mol + 3 * 1g/mol = 17 g/mol

mass in grams = number of moles * molar mass = 189 moles * 17 g/mol = 3213 g

Answer: 3213 g.
</span>
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