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3 years ago

What happens to a container of gas when the pressure is increased?

Chemistry

2 answers:

mrs_skeptik [129]3 years ago

7 0

Answer:

Explanation:When pressure is increased the volume of the gas decreases as Boyle's Law states the the volume of a gas is inversely proportional to pressure.

fiasKO [112]3 years ago

5 0

Answer:

The pressure increases on all surfaces of the container. It begins to heat up. And depending on the strength of the container, it may just break.

Explanation:

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Which is warmer -30C or -30 F?

Burka [1]

-30°C is more warmer than -30°F...!

Always Celsius is warmer than farheniet and also in coldness Celsius is colder than farheneit.

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2 years ago

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13. List the following ions in order of increasing radius: Li+, Mg2+, Br"

Sergeeva-Olga [200]

Answer:

The answer would be B, PC13

Explanation:

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2 years ago

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How many moles are in 4177g of aluminum chloride

Luden [163]

The mole is the unit of measurement in the International System of Units (SI) for amount of substance. It is defined as the amount of a chemical substance that contains as many representative particles, e.g., atoms, molecules, ions, electrons, or photons, as there are atoms in 12 grams of carbon-12 (12C), the isotope of carbon with relative atomic mass 12 by definition.
so to solve the moles, divide the mass with molar mass

moles = 4177 g / 133.34 g/mol
moles = 31.33 moles

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3 years ago

The equilibrium constant Kc for the reaction PCl3(g) + Cl2(g) ⇌ PCl5(g) is 49 at 230°C. If 0.70 mol of PCl3 is added to 0.70 mol

Ymorist [56]

Answer : The correct option is, (B) 0.11 M

Solution :

First we have to calculate the concentration $PCl_3$ and $Cl_2$ .

$\text{Concentration of }PCl_3=\frac{\text{Moles of }PCl_3}{\text{Volume of solution}}$

$\text{Concentration of }PCl_3=\frac{0.70moles}{1.0L}=0.70M$

$\text{Concentration of }Cl_2=\frac{\text{Moles of }Cl_2}{\text{Volume of solution}}$

$\text{Concentration of }Cl_2=\frac{0.70moles}{1.0L}=0.70M$

The given equilibrium reaction is,

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

Initially 0.70 0.70 0

At equilibrium (0.70-x) (0.70-x) x

The expression of $K_c$ will be,

$K_c=\frac{[PCl_5]}{[PCl_3][Cl_2]}$

$K_c=\frac{(x)}{(0.70-x)\times (0.70-x)}$

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

$49=\frac{(x)}{(0.70-x)\times (0.70-x)}$

By solving the term x, we get

$x=0.59\text{ and }0.83$

From the values of 'x' we conclude that, x = 0.83 can not more than initial concentration. So, the value of 'x' which is equal to 0.83 is not consider.

Thus, the concentration of $PCl_3$ at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M

The concentration of $Cl_2$ at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M

The concentration of $PCl_5$ at equilibrium = x = 0.59 M

Therefore, the concentration of $PCl_3$ at equilibrium is 0.11 M

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4 years ago

Glucose, c6h12o6, is used as an energy source by the human body. the overall reaction in the body is described by the equation

Art [367]

Glucose is carbohydrate and a simple sugar that is very important to the human body.

Energy is produced for the cells in the body through the process of metabolism which oxidizes glucose to water, carbon dioxide, and some nitrogen compounds.

The general chemical reaction equation for metabolism is:

C6H12O6 + 6O2 ---> 6CO2 + 6H2O

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4 years ago