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zloy xaker [14]

3 years ago

9

If the volume of wet gas collected over water is 85.0 mL at 20°C and 760 mm Hg , what is the volume of dry gas at STP conditions

? (The vapor pressure of water at 20°C is 17.5 mm Hg.) Express your answer with the appropriate units.

1 answer:

dimulka [17.4K]3 years ago

5 0

Answer: 77.4 mL

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is:

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of dry gas = (760 - 17.5) mmHg= 742.5 mm Hg

$P_2$ = final pressure of dry gas at STP = 760 mm Hg

$V_1$ = initial volume of dry gas = 85.0 mL

$V_2$ = final volume of dry gas at STP = ?

$T_1$ = initial temperature of dry gas = $20^oC=273+20=293K$

$T_2$ = final temperature of dry gas at STP = $0^oC=273+0=273K$

Now put all the given values in the above equation, we get the final volume of wet gas at STP

$\frac{742.5mmHg\times 85.0ml}{293K}=\frac{760mmHg\times V_2}{273K}$

$V_2=77.4mL$

Volume of dry gas at STP is 77.4 mL.

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How many moles are in 110 grams of nahco3

yuradex [85]

Molar mass NaHCO₃ = 23 + 1 + 12 + 16 x 3 = 84 g/mol

1 mole ---------- 84 g
? mole ---------- 110 g

moles NaHCO₃ = 110 . 1 / 84

moles NaHCO₃ = 110 / 84

= 1.309 moles

hope this helps!

4 0

3 years ago

A tank of gas has partial pressures of nitrogen and oxygen equal to 1.61 Ã— 104 kPa and 4.34 Ã— 103 kPa , respectively.What is t

Alchen [17]

Answer:

$20.44\times 10^3 kPa$ is the total pressure of the tank.

Explanation:

Partial pressures of nitrogen = $p_{N_2}=1.61\times 10^4 kPa$

Partial pressure of oxygen = $p_{O_2}=4.34\times 10^3 kPa$

Total pressure of gases in the tank = P

Applying Dalton's law of partial pressures :

$P=p_{N_2}+p_{O_2}=1.61\times 10^4 kPa+4.34\times 10^3 kPa$

$P=20.44\times 10^3 kPa$

$20.44\times 10^3 kPa$ is the total pressure of the tank.

7 0

2 years ago

A bottle containing 415 g of cleaning solution is used to clean hospital equipment. If the cleaning solution has a specific grav

neonofarm [45]

Answer: 483 mL of the cleaning solution are used to clean hospital equipment

Explanation:

The question requires us to calculate the volume, in mL, of solution is used to clean hospital equipment, given that 415g of this solution are used and the specific gravity of the solution is 0.860.

Measurements > Density

Specific gravity is defined as the ratio between the density of a given substance to the density of a reference material, such as water:

$Specific\text{ gravity = }\frac{density\text{ of substance}}{density\text{ of reference substance}}$

The density of a substance is defined as the ratio between the mass and the volume of this substance:

$density=\frac{mass}{volume}$

Considering the reference substance as water and its density as 1.00 g/mL, we can determine the density of the substance which specific gravity is 0.860:

$0.860=\frac{density\text{ of substance}}{1.00g/mL}\rightarrow density\text{ of substance = 0.860g/mL}$

Thus, taking water as the reference substance, we can say that the density of the cleaning solution is 0.860 g/mL.

Now that we know the density of the cleaning solution (0.860 g/mL) and the mass of solution that is used to clean hospital equipment (415g), we can calculate the volume of solution that is used to clean the equipment:

$\begin{gathered} density=\frac{mass}{volume}\rightarrow volume=\frac{mass}{density} \\ \\ volume=\frac{415g}{0.860g/mL}=483mL \end{gathered}$

Therefore, 483 mL of the cleaning solution are used to clean hospital equipment.

6 0

11 months ago

Calculate the percent ionization of formic acid (hco2h) in a solution that is 0.311 m in formic acid and 0.189 m in sodium forma

ohaa [14]

<span>Answer: 0.094%

</span><span>Explanation:
</span>
<span></span><span /><span>
1) Equilibrium chemical equation:
</span><span />

<span>Only the ionization of the formic acid is the important part.
</span><span />

<span>HCOOH(aq) ⇄ HCOO⁻(aq) + H⁺(aq).
</span><span />

<span>2) Mass balance:
</span><span />

<span>                   HCOOH(aq) HCOO⁻(aq) H⁺(aq).

Start 0.311               0.189

Reaction       - x    +x                   +x

Final             0.311 - x    0.189 + x    x

3) Acid constant equation:
</span><span />

<span>Ka = [HCOO-] [N+] / [HCOOH] = (0.189 + x) x / (0.311 -x)
</span><span />

<span>= (0.189 + x )x / (0.311 - x) = 0.000177

4) Solve the equation:

You can solve it exactly (it will lead to a quadratic equation so you can use the quadratiic formula). I suggest to use the fact that x is much much smaller than 0.189 and 0.311.
</span><span />

<span>With that approximation the equation to solve becomes:

</span><span>0.1890x / 0.311 = 0.000177, which leads to:</span>
<span /><span>
x = 0.000177 x 0.311 / 0.189 = 2.91 x 10⁻⁴ M

5) With that number, the percent of ionization (alfa) is:
</span><span />

<span>percent of ionization = (moles ionized / initial moles) x 100 =
</span><span>
</span><span>
</span><span>percent ionization = (concentration of ions / initial concentration) x 100 =
</span><span>
</span><span>
</span><span>percent ionization = (0.000291 / 0.311)x 100 = 0.0936% = 0.094%
</span>
<span></span><span />

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3 years ago

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Aloiza [94]

Answer:

I believe the answer is C

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3 years ago

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