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kykrilka [37]
3 years ago
14

A balloon has a volume of 1.75L at a temperature of 25 °C . what will be the volume of the balloon if you take it out into the w

inter cold air at -15°C?
Chemistry
1 answer:
NemiM [27]3 years ago
3 0
Boyle’s Law P1V1 = P2V2 P1 = 0.80 atm V1 = 1.8 L P2 = 1.0 atm V2 = ?? (.8 atm)(1.8 L) = (1.0 atm)(V2) 1.44 atm x L = 1 atm V2

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The correct answer is A.)

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The correct answer is B.)

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3 years ago
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Diano4ka-milaya [45]
C. Spacecraft are built to be airtight.
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2 years ago
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What is the reaction energy q of this reaction? use c2=931.5mev/u. express your answer in millions of electron volts to three si
Ann [662]
₉₂U²³⁵ + ₀n¹ → ₅₄Xe¹⁴⁰ + ₃₈Sr⁹⁴ + 2 ₀n¹
Mass of reactants = 235.04393 + 1.008665 = 236.052595 amu
Mass of products = 139.92144 + 93.91523 + 2* (1.008665) = 235.854000 amu
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5 0
3 years ago
Determine the pH of 0.050 M HCN solution. HCN is a weak acid with a Ka equal to 4.9 x 10-10<br> DONE
nadezda [96]

Answer:

The pH of the solution is 5.31.

Explanation:

Let "\alpha is the dissociation of weak acid - HCN.

The dissociation reaction of HCN is as follows.

                  HCN+H_{2}O\rightarrow H_{3}O^{+}+CN^{-}

Initial                  C                         0            0

Equilibrium        c(1- \alpha)              c\alpha c\alpha

Dissociation constant = Ka= c\alpha \times \frac{c\alpha}{c(1-\alpha)}

=\frac{c\alpha^{2}}{(1-\alpha)}

In this case weak acids \alpha is very small so, (1-\alpha ) is taken as 1.

Ka=C\alpha^{2}

\alpha=\sqrt\frac{ka}{c}

From the given the concentration = 0.050 M

Substitute the given value.

\alpha=\sqrt\frac{4.9\times 10^{-10}}{0.05}=9.8\times 10^{-4}

[H_{3}O^{+}]=c\alpha

[H_{3}O^{+}]=0.05\times 9.8\times 10^{-4}= 4.9\times10^{-6}

pH= -log[H_{3}O^{+}]

=-log[4.9\times10^{-6}]

=6-log 4.9= 5.31

Therefore, The pH of the solution is 5.31.

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3 years ago
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