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navik [9.2K]
3 years ago
13

Aerosol cans contain a written warning not to subject the can to high temperatures or incinerate. Suppose an aerosol can is heat

ed from 22°C to 110°C. What would you expect to happen? A) The contents of the aerosol can to vaporize. B) No change in volume or pressure; only an increase in temperature. C) The volume of the gas to decreases proportionately to the increase in temperature. D) The pressure in the can to increase proportionately to the increase in temperature.
Chemistry
2 answers:
Zanzabum3 years ago
8 0
D. there would be a proportional increase in pressure to temperature
oksian1 [2.3K]3 years ago
3 0

D, would be your final answer! :)

You might be interested in
Molar mass c3h8
maksim [4K]

Answer:

1. 44.11 g

2. 36.03 g

3. 8.08 g

4. 81.7%

5. 18.3%

Explanation:

1. 12.01+12.01+12.01+1.01+1.01+1.01+1.01+1.01+1.01+1.01+1.01=44.11

2. 12.01×3= 36.03

3. 1.01×8= 8.08

4.(36.03/44.11)×100= 81.7%

5. (8.08/44.11)×100= 18.3%

8 0
2 years ago
What quantity of heat is required to raise the temperature of 460g of aluminum from 15C to 85C?
Hoochie [10]

Answer:

Q = 28.9 kJ

Explanation:

Given that,

Mass of Aluminium, m = 460 g

Initial temperature, T_i=15^{\circ} C

Final temperature, T_f=85^{\circ}

We know that the specific heat of Aluminium is 0.9 J/g°C. The heat required to raise the temperature is given by :

Q=mc\Delta T\\\\Q=mc(T_f-T_i)\\\\Q=460\ g\times 0.9\ J/g^{\circ} C\times (85-15)^{\circ} C\\\\Q=28980\ J\\\\\text{or}\\\\Q=28.9\ kJ

So, 28.9 kJ of heat is required to raise the temperature.

6 0
3 years ago
What is the molarity (M) of the following solutions?
Dennis_Churaev [7]

Answer:

The molarity (M) of the following solutions are :

A. M = 0.88 M

B. M = 0.76 M

Explanation:

A. Molarity (M) of 19.2 g of Al(OH)3 dissolved in water to make 280 mL of solution.

Molar mass of Al(OH)3 = Mass of Al + 3(mass of O + mass of H)

                                      = 27 + 3(16 + 1)

                                      = 27 + 3(17) = 27 + 51

                                      = 78 g/mole

Al(OH)_3 = 78 g/mole

Given mass= 19.2 g/mole

Mole = \frac{Given\ mass}{Molar\ mass}

Mole = \frac{19.2}{78}

Moles = 0.246

Molarity = \frac{Moles\ of\ solute}{Volume\ of\ solution(L)}

Volume = 280 mL = 0.280 L

Molarity = \frac{0.246}{0.280)}

Molarity  = 0.879 M

Molarity  = 0.88 M

B .The molarity (M) of a 2.6 L solution made with 235.9 g of KBr​

Molar mass of KBr = 119 g/mole

Given mass = 235.9 g

Mole = \frac{235.9}{119}

Moles = 1.98

Volume = 2.6 L

Molarity = \frac{Moles\ of\ solute}{Volume\ of\ solution(L)}

Molarity = \frac{1.98}{2.6)}

Molarity = 0.762 M

Molarity = 0.76 M

4 0
2 years ago
A 7.47 g sample of calcium carbonate [CaCO3 (s)] absorbs 85 J of heat, upon which the temperature of the sample increases from 2
Alex_Xolod [135]

Answer:

Specific heat of calcium carbonate(C) = 0.82 (Approx)

Explanation:

Given:

Energy absorbs (q) = 85 J

Change in temperature (Δt) = 34.9 - 21 = 13.9°C  

Mass of calcium carbonate = 7.47 g

Find:

Specific heat of calcium carbonate(C)

Computation:

Specific heat of calcium carbonate(C) = q / m(Δt)

Specific heat of calcium carbonate(C) = 85 / (7.47)(13.9)

Specific heat of calcium carbonate(C) = 85 / 103.833

Specific heat of calcium carbonate(C) = 0.8186

Specific heat of calcium carbonate(C) = 0.82 (Approx)

7 0
2 years ago
1. A 5.04 mL sample of Cu has a mass of 45 grams. What is the density of Cu? Show units in final answer
babunello [35]
I think the answer is number D…. I think
4 0
2 years ago
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