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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! :)

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Which statement is true of a catalyst?
chubhunter [2.5K]

<u><em>Answer:</em></u>

Correct option is D.

It accelerates the reaction rates of a mixture.

<u><em>Explanation:</em></u>

It is used to speed up a reaction by lowering the activation energy.Catalysis is the backbone of many industrial processes, which use chemical reactions to turn raw materials into useful products.

<u><em>Types</em></u>

There are two types of catalyst (1) Homogeneous (2) Heterogeneous

In a heterogeneous reaction, the catalyst is in a different phase from the reactants. In a homogeneous reaction, the catalyst is in the same phase as the reactants.

6 0
4 years ago
When 2.00 g of methane are burned in a bomb calorimeter, the change in temperature is 3.08°C. The heat capacity of the calorimet
melisa1 [442]

Answer:

The approximate molar enthalpy of combustion of this substance is -66 kJ/mole.

Explanation:

First we have to calculate the heat gained by the calorimeter.

q=c\times \Delta T

where,

q = Heat gained = ?

c = Specific heat = 2.68 kJ/^oC

ΔT =  The change in temperature = 3.08°C

Now put all the given values in the above formula, we get:

q=2.68 kJ/^oC\times 3.08^oC

q=8.2544 kJ

Now we have to calculate molar enthalpy of combustion of this substance :

\Delta H_{comb}=-\frac{q}{n}

where,

\Delta H_{comb} = enthalpy change = ?

q = heat gained = 8.2544kJ

n = number of moles methane = \frac{\text{Mass of methane}}{\text{Molar mass of methane }}=\frac{2.00 g}{16.042 g/mol}=0.1247 mole

\Delta H_{comb}=-\frac{8.2544 kJ}{0.1247 mole}=-66.21 kJ/mole\approx -66 kJ/mole

Therefore,  the approximate molar enthalpy of combustion of this substance is -66 kJ/mole.

6 0
3 years ago
I NEED HELP ASAP PLZ LOOK AT THE PICTURE CAREFULLY!!!!
Virty [35]

Answer:

it is a little fuzzy

Explanation:

6 0
3 years ago
Determine el PH y el % de disociación de una solución de ácido débil, sabiendo que se disuelven 20 gramos del ácido (masa molar=
IceJOKER [234]

Answer:

pH = 4.27. Porcentaje de disociación: 0.03%

Explanation:

El pH de un ácido débil, HX, se obtiene haciendo uso de su equilibrio:

HX(aq) ⇄ H⁺(aq) + X⁻(aq)

Donde la constante de equilibrio, Ka, es

Ka = 1.65x10⁻⁸ = [H⁺] [X⁻] / [HX]

Como los iones H⁺ y X⁻ vienen del mismo equilibrio podemos decir:

[H⁺] = [X⁻]

[HX] es:

20g * (1mol/55g) = 0.3636moles / 2.100L = 0.1732M

Reemplazando es Ka:

1.65x10⁻⁸ = [H⁺] [H⁺] / [0.1732M]

2.858x10⁻⁹ = [H⁺]²

5.35x10⁻⁵M = [H⁺]

pH = -log[H⁺]

<h3>pH = 4.27</h3>

El porcentaje de disociacion es [X⁻] / [HX] inicial * 100

Reemplazando

5.35x10⁻⁵M / 0.1732M * 100

<h3>0.03%</h3>
5 0
3 years ago
A change in the water temperature of the pacific ocean that produce warm current
liraira [26]
This is true, this isn't a question, it's a fact.
6 0
3 years ago
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