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nika2105 [10]
2 years ago
13

What is the pressure of 0.33 moles of nitrogen gas, if its volume is 15.0 L at –25.0oC?

Chemistry
1 answer:
musickatia [10]2 years ago
3 0

Using the ideal gas law PV =nRTPV=nRT , we find that the pressure will be P =\frac{nRT}{V}P=

V

nRT

​

 . Then, we'll substitute and find the pressure, using T = -25 °C = 248.15 K and R = 0.0821 \frac{atm\cdot L}{mol \cdot K}

mol⋅K

atm⋅L

​

 :

P =\frac{nRT}{V} = \frac{(0.33\,\cancel{mol})(0.0821\frac{atm\cdot \cancel{L}}{\cancel{mol \cdot K}})(248.15\,\cancel{K})}{15.0\,\cancel{L}} = 0.4482\,atmP=

V

nRT

​

=

15.0

L

​

(0.33

mol

)(0.0821

mol⋅K

atm⋅

L

​

​

)(248.15

K

​

)

​

=0.4482atm

In conclusion, the pressure of this gas is P=0.4482 atm.

Reference:

Chang, R. (2010). Chemistry. McGraw-Hill, New York.

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Answer:

CuSO4 + 2 NaOH = Cu(OH)2 + Na2SO4

3 0
3 years ago
The expected value for a chemical equation is 47g of water, after an experiment you find that you have 2.58 moles of water. What
strojnjashka [21]
<h3><u>Answer</u>;</h3>

Actual yield = 46.44 g

<h3><u>Explanation;</u></h3>

1 mole of water = 18 g/mol

Therefore;

The experimental yield = 2.58 moles

equivalent to ; 2.58 × 18 = 46.44 g

The theoretical value is 47 g

Percentage yield = 46.44/47 × 100%

                             = 98.8%

The questions asks for actual yield = 46.44 g

6 0
3 years ago
A wooden artifact from an ancient tomb contains 60% of the carbon-14 that is present in living trees. How long ago was the artif
noname [10]

Answer:

The age of the sample is 4224 years.

Explanation:

Let the age of the sample be t years old.

Initial mass percentage of carbon-14 in an artifact = 100%

Initial mass of carbon-14 in an artifact = [A_o]

Final mass percentage of carbon-14 in an artifact t years = 60%

Final mass of carbon-14 in an artifact = [A]=0.06[A_o]

Half life of the carbon-14 = t_{1/2}=5730 years

k=\frac{0.693}{t_{1/2}}

[A]=[A_o]\times e^{-kt}

[A]=[A_o]\times e^{-\frac{0.693}{t_{1/2}}\times t}

0.60[A_o]=[A_o]\times e^{-\frac{0.693}{5730 year}\times t}

Solving for t:

t = 4223.71 years ≈ 4224 years

The age of the sample is 4224 years.

8 0
3 years ago
how many mL of a 3.5 M sodium hydroxide will be needed to neutralize 15mL of a 4.3 M hydrochloric acid solution?
kobusy [5.1K]
Step 1 : write a valanced equation..
NaOH + HCl 》NaCl + H2O

Step 2 : find the number of mole of HCl..
1000 ml ..contains 4.3 mole
15ml... (4.3÷1000)×15 =...

Stem 3 : use mole ratio....
HCl : NaOH
1 : 1
So mole is same as calculated above...

Step 4 :
3.5 mole of NaOH is in 1000ml
(4.3÷1000)×15 mole is in ....


Do the calculation
6 0
3 years ago
a 25.5 liter balloon holding 3.5 moles of carbon dioxide leaks. If we are able to determine that 1.9 moles of carbon dioxide esc
kap26 [50]

Answer:

11.66 L.

Explanation:

  • We can use the general law of ideal gas: <em>PV = nRT.</em>

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

  • If P and T are constant, and have different values of n and V:

<em>(V₁n₂) = (V₂n₁).</em>

V₁ = 25.5 L, n₁ = 3.5 mol.

V₂ = ??? L, n₂ = 3.5 mol - 1.9 mol = 1.6 mol.

<em>∴ V₂ = (V₁n₂)/(n₁)</em> = (25.5 L)(1.6 mol)/(3.5 mol) =<em> 11.66 L.</em>

4 0
3 years ago
Read 2 more answers
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