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kondor19780726 [428]
1 year ago
6

Suppose that on a hot and sticky afternoon in the spring, a tornado passes over the high school. If the air pressure in the lab

(volume of 180 m3) was 1.1 atm before the storm and 0.85 atm during the storm, to what volume would the laboratory try to expand in order to make up for the large pressure difference outside?
190 m3
230 m3
1,800 m3
7,100 m3
Chemistry
2 answers:
uranmaximum [27]1 year ago
5 0

The answer is B) 230 m3

Andreyy891 year ago
5 0

Answer: 230m^3

Explanation:

Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}{V}     (At constant temperature and number of moles)

P_1V_1=P_2V_2

where,

P_1 = initial pressure of gas  = 1.1 atm

P_2 = final pressure of gas  = 0.85 atm

V_1 = initial volume of gas  = 180m^3

V_2 = final volume of gas  = ?

1.1\times 180=0.85\times V_2

V_2=230m^3

Therefore, the gas would expand to volume of 230m^3

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Explanation:

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If 1.50 μg of co and 6.80 μg of h2 were added to a reaction vessel, and the reaction went to completion, how many gas particles
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Mrs. Rushing fills a balloon with hydrogen gas to demonstrate its ability to burn. Which combination could she
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A 0.100 m solution of which one of the following solutes will have the highest vapor pressure? A 0.100 m solution of which one o
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Explanation:

Vapor pressure is defined as the pressure exerted by vapors or gas on the surface of a liquid.

Vapor pressure is inversely proportional to the number of solute particles. Hence, more will be the solute particles lower will be the vapor pressure and vice-versa.

(a)   KClO_{4} \rightarrow K^{+} + ClO^{-}_{4}

It dissociates to give two particles.

(b)  Ca(ClO_{4})_{2} \rightarrow Ca^{2+} + 2ClO^{-}_{4}

Total number of particles it give upon dissociation are 1 + 2 = 3. Hence, it gives 3 particles.

(c)   Al(ClO_{4})_{3} \rightarrow Al^{3+} + 3ClO^{-}_{4}

Total number of particles it give upon dissociation are 1 + 3 = 4. Hence, it gives 4 particles.

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2 years ago
A 600.0 mL sample of 0.20 MHF is titrated with 0.10 MNaOH. Determine the pH of the solution after the addition of 600.0 mL of Na
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Answer: pH=12.69

Explanation:

{\text{Moles of HF}=Molarity\times {\text{Volume of solution in liters}}

{\text{Moles of HF}=0.20M\times 0.6L=0.12 moles

HF\rightarrow H^++F^-

Initial 0.12               0       0

Eqm   0.12-x           x        x

K_a=\frac{[H^+][F^-]}{[HF]}

3.5\times 10^{-4}=\frac{x^2}{0.12-x}  

(neglecting small value of x in comparison to 0.12)

x=4.2\times 10^{-5}

Moles of H^+=4.2\times 10^{-5}

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{\text{Molesof NaOH}}=Molarity\times {\text{Volume of solution in liters}}

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0.06 moles of NaOH will give 0.06 moles of [OH^-]

Now 4.2\times 10^{-5} moles of OH^- will be neutralized by 4.2\times 10^{-5} moles of H^+ and (0.06-4.2\times 10^{-5})=0.059 moles of OH^- will be left.

Molarity of OH^-=\frac{0.059moles}{1.2L}=0.049M

pOH=-\log[OH^-]=-\log[0.049]=1.31

pH = 14 - pOH= 14 - 1.31 = 12.69

5 0
2 years ago
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