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alexira [117]
2 years ago
15

In which example is matter changing

Chemistry
1 answer:
just olya [345]2 years ago
6 0
A) heating a pan of water until the water is all gone because then it would change from a liquid top a gas.
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What volume does 0.482 mol of gas occupy at a pressure of 719 mmHg at 56 ∘C?
kozerog [31]

Answer:

The volume is 13, 69 L

Explanation:

We use the formula PV=nRT. We convert the temperature in Celsius into Kelvin and the pressure in mmHg into atm.

0°C= 273K---> 56°C= 56 + 273= 329K

760 mmHg----1 atm

719 mmHg----x= (719 mmHgx 1 atm)/760 mmHg= 0,95 atm

PV=nRT ---> V= (nRT)/P

V=( 0,482 molx 0,082 l atm/K mol x 329K)/0,95 atm

<em>V=13,68778526 L</em>

7 0
3 years ago
Br2(g) cl2(g)⇌2brcl(g) δh∘f for brcl(g) is 14. 6 kj/mol. Δs∘f for brcl(g) is 240. 0 j/mol
max2010maxim [7]

The Change in Gibb's free energy, ΔG for the reaction at 298K is; -56.92KJ.

<h3>Gibb's free energy of reactions</h3>

It follows from the Gibb's free energy formula as expressed in terms of Enthalpy and Entropy that;

  • ΔG = ΔH - TΔS

On this note, it follows that;

  • ΔG = 14.6 - (298× 0.24)

Hence, the Gibb's free energy for the reaction is;

  • ΔG = 14.6 - 71.52
  • ΔG = -56.92KJ

Remarks: The question requires that we determine the Gibb's free energy for the reaction at 298K.

Read more on Gibb's free energy;

brainly.com/question/13765848

5 0
1 year ago
Read 2 more answers
Given that the freezing point depression constant for water is 1.86°c kg/mol, calculate the change in freezing point for a 0.907
melamori03 [73]

Answer : The correct answer for change in freezing point = 1.69 ° C

Freezing point depression :

It is defined as depression in freezing point of solvent when volatile or non volatile solute is added .

SO when any solute is added freezing point of solution is less than freezing point of pure solvent . This depression in freezing point is directly proportional to molal concentration of solute .

It can be expressed as :

ΔTf = Freezing point of pure solvent - freezing point of solution = i* kf * m

Where : ΔTf = change in freezing point (°C)

i = Von't Hoff factor

kf =molal freezing point depression constant of solvent.\frac{^0 C}{m}

m = molality of solute (m or \frac{mol}{Kg} )

Given : kf = 1.86 \frac{^0 C*Kg}{mol}

m = 0.907 \frac{mol}{Kg} )

Von't Hoff factor for non volatile solute is always = 1 .Since the sugar is non volatile solute , so i = 1

Plugging value in expression :

ΔTf = 1* 1.86 \frac{^0 C*Kg}{mol} * 0.907\frac{mol}{Kg} )

ΔTf = 1.69 ° C

Hence change in freezing point = 1.69 °C

5 0
3 years ago
10. Well into the twentieth century, many scientists believed the ocean floor was A. rocky. B. round. C. flat. D. sandy.
Murljashka [212]
<span>In the 20th century, the scientist believed that the floor of the ocean floor is flat. But later on discovered that their theory is wrong. The ocean floor is not flat and featureless. One technology helped the scientist to study in advance and in details the real feature of the ocean floor. They used sound waves to know the distance of the water from floor level to the top. They also discovered that ranges of mountains lies under the ocean. Sonar was one of the technologies they used to study the sound waves emitted by the ocean. Hope i was helpful :)</span>
8 0
3 years ago
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What is the acceleration of a jogger who moves in a straight line and increases her speed by 1 mile per hour every 0.5 hours?
Reil [10]
2 miles per every hour ?
6 0
2 years ago
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