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4 years ago

Which of the following can be characterized as a physical change?

Chemistry

1 answer:

Taya2010 [7]4 years ago

7 0

Answer:

D. Sweat evaporates from your skin

Explanation:

A physical change is a reversible process, no new substance is formed. A physical change does not affect the chemical composition of a substance.

Sweat evaporation from the skin is a physical change, because the change that occur is merely a change of state, Liquid to Gaseous state, no new substance is formed.

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1.00 L of a gas at STP is compressed to 473mL. What is the new pressure of gas?

Ratling [72]

Hello!

1.00 L of a gas at STP is compressed to 473 mL. What is the new pressure of gas?

We have the following data:

Vo (initial volume) = 1.00 L

V (final volume) = 473 mL → 0.473 L

Po (initial pressure) = 1 atm (pressure exerted by the atmosphere - in STP)

P (final pressure) = ? (in atm)

We have an isothermal transformation, that is, its temperature remains constant, if the volume of the gas in the container decreases, so its pressure increases. Applying the data to the equation Boyle-Mariotte, we have:

$P_0*V_0 = P*V$

$1*1 = P*0.473$

$1 = 0.473\:P$

$0.473\:P = 1$

$P = \dfrac{1}{0.473}$

$\boxed{\boxed{P \approx 2.11\:atm}}\:\:\:\:\:\:\bf\green{\checkmark}$

Answer:

The new pressure of the gas is 2.11 atm

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$\bf\blue{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}$

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3 years ago

The form of heat transfer taking place in the crust of the earth would be ____

Anna [14]

The form of heat transfer taking place in the crust of the Earth would be Convection.

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Write the IUPAC names for the following compounds.

atroni [7]

Benzine
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2 years ago

Calculate the freezing point of the solution containing 0.105 m k2s. calculate the boiling point of the solution above. -g

Licemer1 [7]

Quantity of K2S m = 0.105 m
Number of ions i = 2(K) + 1(S) = 3
Freezing point depression constant of water Kf = 1.86
delta T = i x m x Kf = 3 x 0.105 x 1.86 = 0.586
Freezing point = 0 - 0.586 = 0.586 C
Boiling point constant of water Kb = 0.512
delta T = i x m x Kb = 3 x 0.105 x 0.512 = 0.161
Boiling point = 100 + 0.161 = 100.161 C

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3 years ago

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3 years ago