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Gennadij [26K]
3 years ago
5

During laparoscopic surgery , carbon dioxide gas is used to expand the abdomen to help create a larger working space. If 4.80 L

of CO 2 gas at 18 °C at 785 mmHg is used, what is the final volume , in liters, of the gas at 37 degrees * C and a pressure of 745 mmHg if the amount of CO 2 does not change ?
Chemistry
1 answer:
Studentka2010 [4]3 years ago
7 0

Answer:

5.37 L

Explanation:

To solve this problem we need to use the PV=nRT equation.

First we <u>calculate the amount of CO₂</u>, using the initial given conditions for P, V and T:

  • P = 785 mmHg ⇒ 785/760 = 1.03 atm
  • V = 4.80 L
  • T = 18 °C ⇒ 18 + 273.16 = 291.16 K

1.03 atm * 4.80 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 291.16 K

We <u>solve for n</u>:

  • n = 0.207 mol

Then we use that value of n for another PV=nRT equation, where T=37 °C (310.16K) and P = 745 mmHg (0.98 atm).

  • 0.98 atm * V = 0.207 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 310.16 K

And we <u>solve for V</u>:

  • V = 5.37 L
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What is the pOH of a 0.150 M solution of potassium nitrite? (Ka HNO2 = 4.5 x 10−4 )
yanalaym [24]

Answer:

11.9 is the pOH of a 0.150 M solution of potassium nitrite.

Explanation:

Solution :  Given,

Concentration (c) = 0.150 M

Acid dissociation constant = k_a=4.5\times 10^{-4}

The equilibrium reaction for dissociation of HNO_2 (weak acid) is,

                           HNO_2+H_2O\rightleftharpoons NO_2^-+H_3O^+

initially conc.         c                       0         0

At eqm.              c(1-\alpha)                c\alpha        c\alpha

First we have to calculate the concentration of value of dissociation constant (\alpha ).

Formula used :

k_a=\frac{(c\alpha)(c\alpha)}{c(1-\alpha)}

Now put all the given values in this formula ,we get the value of dissociation constant (\alpha ).

4.5\times 10^{-4}=\frac{(0.150\alpha)(0.150\alpha)}{0.150(1-\alpha)}

4.5\times 10^{-4} - 4.5\times 10^{-4}\alpha =0.150\alpha ^2

0.150\alpha ^2+4.5\times 10^{-4}\alpha-4.5\times 10^{-4}=0

By solving the terms, we get

\alpha=0.0533

No we have to calculate the concentration of hydronium ion or hydrogen ion.

[H^+]=c\alpha=0.150\times 0.0533=0.007995 M

Now we have to calculate the pH.

pH=-\log [H^+]

pH=-\log (0.007995 M)

pH=2.097\approx 2.1

pH + pOH = 14

pOH =14 -2.1 = 11.9

Therefore, the pOH of the solution is 11.9

4 0
3 years ago
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