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3 years ago

10

If temperature is held constant for an ideal gas, as P and V change for a given gas sample, the PV of products will_____. increa

se decrease be constant change

2 answers:

noname [10]3 years ago

6 0

Answer:

The PV of products will be constant.

Explanation:

Ideal gas equation- PV = nRT

where P is pressure of gas, V is volume of gas, n is number of moles of gas, R is gas constant and T is temperature (in Kelvin) of gas.

Now, number of moles of a gas remain constant with change in state variables.

So, if T is constant then the term "nRT" remins constant.

As PV is equal to nRT therefore PV will be constant if temperature is held constant.

In another way we can say if P is increased then V will decrease or vice versa.

vodka [1.7K]3 years ago

3 0

Answer:

PV=nRT

Explanation:

V=<u>R</u><u>T</u><u>n</u>

P

rearrangement gives

PV=nRT
R=<u>P</u><u>V</u>

nT

where P=pressure

V=volume

n=number of moles

R=ideal gas(0.0820atmdm/3 mol/k)

T=temperature in kelvin

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Calculate the percent of each component in the mixture. Show your calculations. Circle final answers.

Colt1911 [192]

Answer:

See Explanation

Explanation:

The question is incomplete; as the mixtures are not given.

However, I'll give a general explanation on how to go about it and I'll also give an example.

The percentage of a component in a mixture is calculated as:

$\%C_E = \frac{E}{T} * 100\%$

Where

E = Amount of element/component

T = Amount of all elements/components

Take for instance:

In $(Ca(OH)_2)$

The amount of all elements is: (i.e formula mass of $(Ca(OH)_2)$ )

$T = 1 * Ca + 2 * H + 2 * O$

$T = 1 * 40 + 2 * 1 + 2 * 16$

$T = 74$

The amount of calcium is: (i.e formula mass of calcium)

$E = 1 * Ca$

$E = 1 * 40$

$E = 40$

So, the percentage component of calcium is:

$\%C_E = \frac{E}{T} * 100\%$

$\%C_E = \frac{40}{74} * 100\%$

$\%C_E = \frac{4000}{74}\%$

$\%C_E = 54.05\%$

The amount of hydrogen is:

$E = 2 * H$

$E = 2 * 1$

$E = 2$

So, the percentage component of hydrogen is:

$\%C_E = \frac{E}{T} * 100\%$

$\%C_E = \frac{2}{74} * 100\%$

$\%C_E = \frac{200}{74}\%$

$\%C_E = 2.70\%$

Similarly, for oxygen:

The amount of oxygen is:

$E = 2 * O$

$E = 2 * 16$

$E = 32$

So, the percentage component of oxygen is:

$\%C_E = \frac{E}{T} * 100\%$

$\%C_E = \frac{32}{74} * 100\%$

$\%C_E = \frac{3200}{74}\%$

$\%C_E = 43.24\%$

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A sample of gas with an initial volume of 12.5 L at a pressure of 784 torr and a temperature of 295 K is compressed to a volume

Kipish [7]

Answer:

Final pressure in (atm) (P1) = 6.642 atm

Explanation:

Given:

Initial volume of gas (V) = 12.5 L

Pressure (P) = 784 torr

Temperature (T) = 295 K

Final volume (V1) = 2.04 L

Final temperature (T1) = 310 K

Find:

Final pressure in (atm) (P1) = ?

Computation:

According to combine gas law method:

$\frac{PV}{T} =\frac{P1V1}{T1} \\\\\frac{(784)(12.5)}{295} =\frac{(P1)(2.04)}{310}\\\\33.22 = \frac{(P1)(2.04)}{310}\\\\P1=5,048.18877$

⇒ Final pressure (P1) = 5,048.18877 torr

⇒ Final pressure in (atm) (P1) = 5,048.18877 torr / 760

⇒ Final pressure in (atm) (P1) = 6.642 atm

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vaieri [72.5K]

Answer:

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Explanation:

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anyanavicka [17]

This problem is providing information about the initial mass of mercury (II) oxide (10.00 g) which is able to produce liquid mercury (8.00 g) and gaseous oxygen and asks for the resulting mass of the latter, which turns out to be 0.65 g after doing the corresponding calculations.

Initially, it is given a mass of 10.00 g of the oxide and 1.35 g are left which means that the following mass is consumed:

$m_{HgO}^{consumed}=10.00g-1.35 g=8.65 g$

Now, since 8.00 grams of liquid mercury are collected, it is possible to calculate the grams of oxygen that were produced, by considering the law of conservation of mass, which states that the mass of the products equal that of the reactants as it is nor destroyed nor created. In such a way, the mass of oxygen turns out to be:

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Which use of iron is due to its chemical properties?

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Explanation:

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