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3 years ago

What causes an ice cube to melt when removed from a freezer?

Chemistry

2 answers:

OverLord2011 [107]3 years ago

4 0

Answer:

the melting process begins right away because the air temperature around the ice cubes is warmer than the temperature in the freezer

iren2701 [21]3 years ago

3 0

The temperature in the room is higher than the temperature in the freezer, thus causing the melting state of the ice cube when it is removed from the freezer.

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A gas mixture with a total pressure of 745 mmHg contains each of the following gases at the indicated partial pressures: CO2, 24

Setler79 [48]

Answer:

For Part A: The partial pressure of Helium is 218 mmHg.

For Part B: The mass of helium gas is 0.504 g.

Explanation:

For Part A:

We are given:

$p_{CO_2}=245mmHg\\p_Ar}=119mmHg\\p_{O_2}=163mmHg\\P=745mmHg$

To calculate the partial pressure of helium, we use the formula:

$P=p_{CO_2}+p_{Ar}+p_{O_2}+p_{He}$

Putting values in above equation, we get:

$745=245+119+163+p_{He}\\p_{He}=218mmHg$

Hence, the partial pressure of Helium is 218 mmHg.

For Part B:

To calculate the mass of helium gas, we use the equation given by ideal gas:

PV = nRT

or,

$PV=\frac{m}{M}RT$

where,

P = Pressure of helium gas = 218 mmHg

V = Volume of the helium gas = 10.2 L

m = Mass of helium gas = ? g

M = Molar mass of helium gas = 4 g/mol

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

T = Temperature of helium gas = 283 K

Putting values in above equation, we get:

$218mmHg\times 10.2L=\frac{m}{4g/mol}\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 283K\\\\m=0.504g$

Hence, the mass of helium gas is 0.504 g.

6 0

3 years ago

You measure the mass of a model car to be 230 grams.The actual mass is 218 grams .what is your percent error

LenKa [72]

Your percent is 5.50458716

8 0

3 years ago

Our day is 24 hours long because that is the time it takes the earth to

Marianna [84]

Complete one rotation.

Hope this helps.

3 0

3 years ago

Read 2 more answers

I need help with 4, 5, 8, 9, and 6. Quickly I need it before class starts. Worth points!!!!!! HelP

GalinKa [24]

Answer:

4. 264.6J

5. 37.5J

6. 96J

7. 55Watts

8. 77.14m

9. 6s

10. 750Watts

Explanation:

4). Mechanical energy (potential energy) = mass (m) × acceleration due to gravity (g) × height (h)

m = 3kg, h = 9m, g = 9.8m/s²

P.E = 3 × 9 × 9.8

= 264.6J

5). Kinetic energy (K.E) = 1/2 × m × v²

Where;

m = mass (kg) = 3kg

v = velocity (m/s) = 5m/s

K.E = 1/2 × 3 × 5²

K.E = 1/2 × 3 × 25

K.E = 1/2 × 75

K.E = 37.5J

6). Work done (J) = Force (N) × distance (m)

Force = 12N, distance = 8m

Work done = 12 × 8

= 96J

7). Power = work done (J) ÷ time (s)

Work done = 550J, time = 10s

Power = 550/10

= 55Watts.

8). Work done = force (F) × distance (m)

Work done = 540J, force = 7N, distance = ?

540 = 7 × d

540 = 7d

d = 540/7

d = 77.14m

9). Power = work done (J) ÷ time (s)

Work done = 300J, time = ?, Power = 50Watts.

50 = 300/t

50t = 300

t = 300/50

t = 6seconds.

10). Power = work done (J) ÷ time (s)

This means that;

Power = force × distance / time

Force = 300N, distance = 5m, time = 2s

Power = 300 × 5 ÷ 2

Power = 1500 ÷ 2

Power = 750Watts

3 0

3 years ago

If the initial [NO2] is 0.260 M, it will take ________ s for the concentration to drop to 0.150 M. If the initial is 0.260 , it

Nitella [24]

The question is incomplete, here is the complete question:

At elevated temperature, nitrogen dioxide decomposes to nitrogen oxide and oxygen gas

$NO_2\rightarrow NO+\frac{1}{2}O_2$

The reaction is second order for $NO_2$ with a rate constant of $0.543M^{-1}s^{-1}$ at 300°C. If the initial [NO₂] is 0.260 M, it will take ________ s for the concentration to drop to 0.150 M