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Roman55 [17]
3 years ago
15

which equation is setup correctly to determine the volume of a 1.5 mol sample of oxygen gas at 22 degrees Celsius and 100 kPa

Chemistry
1 answer:
rjkz [21]3 years ago
4 0
Hello!

We have the following data:


v (volume) = ? (in L)
n (number of mols) = 1,5 mol
T (temperature) = 22 ºC 
First let's convert the temperature on the Kelvin scale, let's see:
TK = TºC + 273,15
TK = 22 + 273,15
TK = 295,15

P (pressure) = 100 kPa → P = 100000 Pa → P ≈ 0,987 atm
R (gas constant) = 0,082 atm.L / mol.K

<span>We apply the data above to the Clapeyron equation (gas equation), let's see:

</span>P*V = n*R*T

0,987*V = 1,5*0,082*295,15

0,987V = 36,30345

V =  \dfrac{36,30345}{0,987}

\boxed{\boxed{V \approx 36,78\:L}} \end{array}}\qquad\checkmark

I hope this helps. =)
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A 10.0 mL sample of 0.25 M NaOH(aq) is titrated with 0.10 M HCl(aq) (adding HCl to NaOH). Determine which region on the titratio
Anna11 [10]

Answer:

1) After adding 15.0 mL of the HCl solution, the mixture is before the equivalence point on the titration curve.

2) The pH of the solution after adding HCl is 12.6

Explanation:

10.0 mL of 0.25 M NaOH(aq) react with 15.0 mL of 0.10 M HCl(aq). Let's calculate the moles of each reactant.

nNaOH=\frac{0.25mol}{L} .10.0 \times 10^{-3} L=2.5 \times 10^{-3}mol

nHCl=\frac{0.10mol}{L} \times 15.0 \times 10^{-3} L=1.5 \times 10^{-3}mol

There is an excess of NaOH so the mixture is before the equivalence point. When HCl completely reacts, we can calculate the moles in excess of NaOH.

                    NaOH       +       HCl       ⇒       NaCl      +         H₂O

Initial          2.5 × 10⁻³         1.5 × 10⁻³               0                      0

Reaction    -1.5 × 10⁻³        -1.5 × 10⁻³          1.5 × 10⁻³          1.5 × 10⁻³

Final            1.0 × 10⁻³               0                 1.5 × 10⁻³          1.5 × 10⁻³

The concentration of NaOH is:

[NaOH]=\frac{1.0 \times 10^{-3} mol }{25.0 \times 10^{-3} L} =0.040M

NaOH is a strong base so [OH⁻] = [NaOH].

Finally, we can calculate pOH and pH.

pOH = -log [OH⁻] = -log 0.040 = 1.4

pH = 14 - pOH = 14 - 1.4 = 12.6

5 0
3 years ago
A biology student set up an experiment to study 9 mice. On day 1, they were measured for weight and length. Then the mice were p
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The independent variable would be the variable in the research that is being manipulated by the researcher. In this case, it would be temperature in the cage as it is what is being manipulated and changed in the research design. The dependent variable would be the variable that is being studied so, for this case, it would be the length and the weight of the mice. The constants are the factors that might affect the dependent variable but is held constant or the same by the researcher throughout the experiment. These are the size of the cage, amount of food and the exercise wheel. The flaw that the scientist would be studying the length of the mice since I don't think the temperature has any effect on it. And base from he results, the change in lengths are not conclusive.
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A naturally occurring oil co-distills with water to produce an oil/water distillate that is 20% oil by weight. If the molecular
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Answer:

Explanation:

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= mole of oil = mass of oil / molecular weight of oil

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