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FrozenT [24]
3 years ago
15

What is the total pressure in atmospheres in a 10.0L vessel containing 2.50 x 10-3 mol H2, 1.00 x 10-3

Chemistry
1 answer:
grigory [225]3 years ago
8 0

Answer:

Total pressure = 16.42× 10⁻⁹atm

Explanation:

Given data:

Moles of H₂ = 2.50 × 10⁻³ mol

Moles of He = 1.00 × 10⁻³ mol

Mass of Ne = 3 × 10⁻⁴ mol

Volume = 10 L

Temperature = 35°C

Total pressure = ?

Solution:

Pressure of hydrogen:

P = nRT / V

P = 2.50 × 10⁻³ mol× 0.0821 atm. L.mol⁻¹ .k⁻¹ × 308 K / 10 L

p = 63.22× 10⁻³  atm. L /10 L

P = 6.3 × 10⁻³atm

Pressure of helium:

P = nRT / V

P = 1.00 × 10⁻³ mol× 0.0821 atm. L.mol⁻¹ .k⁻¹ × 308 K / 10 L

p = 25.29 × 10⁻³ atm. L /10 L

P = 2.53× 10⁻³ atm

Pressure of neon:

P = nRT / V

P = 3 × 10⁻⁴ mol× 0.0821 atm. L.mol⁻¹ .k⁻¹ × 308 K / 10 L

p = 75.86× 10⁻³ atm. L /10 L

P = 7.59× 10⁻³ atm

Total pressure of mixture:

P(mixture)  = pressure of hydrogen + pressure of helium+ pressure of neon

P(mixture)  = 6.3 × 10⁻³atm + 2.53× 10⁻³ atm + 7.59× 10⁻³ atm

P(mixture) = 16.42× 10⁻⁹atm

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How many grams of o2 are required to produce 358.5 grams of zno? 2zn + o2 ® 2zno?
Murrr4er [49]
The balanced equation for the reaction is ;
2Zn + O2 —> 2ZnO
The stoichiometry of O2 to ZnO is 1:2
The mass of ZnO formed - 358.5 g
The number of moles formed - 358.5 g / 81.4 g/mol = 4.4 moles
Therefore number of O2 moles reacted = 4.4 moles /2 = 2.2 mol
Mass of O2 reacted = 2.2 mol x 32 g/mol = 70.4 g
6 0
3 years ago
A sample of oxygen gas occupies a volume of 160 liters at 364 K. What will be the volume of the gas when the temperature drops t
Ivahew [28]

Answer:

62L

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tysm.

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hope it help

8 0
3 years ago
When 5.00 g of Al2S3 and 2.50 g of H2O are reacted according to the following reaction: Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H
Debora [2.8K]

Answer:

Y=58.15\%

Explanation:

Hello,

For the given chemical reaction:

Al_2S_3(s) + 6 H_2O(l) \rightarrow 2 Al(OH)_3(s) + 3 H_2S(g)

We first must identify the limiting reactant by computing the reacting moles of Al2S3:

n_{Al_2S_3}=5.00gAl_2S_3*\frac{1molAl_2S_3}{150.158 gAl_2S_3} =0.0333molAl_2S_3

Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:

n_{Al_2S_3}^{consumed}=2.50gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molAl_2S_3}{6molH_2O}=0.0231mol  Al_2S_3

Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:

m_{Al(OH)_3}=0.0231molAl_2S_3*\frac{2molAl(OH)_3}{1molAl_2S_3}*\frac{78gAl(OH)_3}{1molAl(OH)_3} =3.61gAl(OH)_3

Finally, we compute the percent yield with the obtained 2.10 g:

Y=\frac{2.10g}{3.61g} *100\%\\\\Y=58.15\%

Best regards.

7 0
3 years ago
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pshichka [43]

Answer:???????????????????????? huh?

Explanation:

6 0
3 years ago
Please help in Chemistry!!!! 15 points! I added a picture below...
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The answer would be choice B because the energy decreased by 20 J
5 0
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