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3 years ago

If the pressure of a sample of gas is doubled at constant temperature, what happens to the volume of the gas?

Chemistry

2 answers:

Sloan [31]3 years ago

5 0

I just took the test and the correct answer is B.) the volume is halved

patriot [66]3 years ago

4 0

The volume gas will stay constant

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A 3.31-g sample of lead nitrate, Pb(NO3)2, molar mass = 331 g/mol, is heated in an evacuated cylinder with a volume of 2.53 L. T

Dahasolnce [82]

Answer:

0.486atm is the pressure of the cylinder

Explanation:

1 mole of Pb(NO₃)₂ descomposes in 4 moles of NO2 and 1 mole of O2. That is 1 mole descomposes in 5 moles of gas.

To find the pressure of the cylinder, we need to find moles of gas produced, and using general gas law we can determine the pressure of the gas:

Moles Pb(NO₃)₂ and moles of gas:

3.31g * (1mol / 331g) = 0.01 moles of Pb(NO₃)₂.

That means moles of gas produced is 0.05 moles.

Pressure of the gas:

Using PV = nRT

P = nRT/V

Where P is pressure (Incognite)

V is volume (2.53L)

R is gas constant (0.082atmL/molK)

T is absolute temperature (300K)

And n are moles of gas (0.05 moles)

P = 0.05mol*0.082atmL/molK*300K / 2.53L

P = 0.486atm is the pressure of the cylinder

3 0

3 years ago

Which observation does not indicate that a chemical reaction has occurred?

castortr0y [4]

Answer:

change in the total mass of substances

6 0

3 years ago

30cm^3 of a dilute solution of Ca(OH)2 required 11 cm^3 of 0.06 mol/dm^. Hcl for complete neutralization. Calculate the concentr

Alenkasestr [34]

Answer: Thus concentration of $Ca(OH)_2$ in $mol/dm^3$ is 0.011 and in $g/dm^3$ is 0.814

Explanation:

To calculate the concentration of $Ca(OH)_2$ , we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HCl$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is $Ca(OH)_2$

We are given:

$n_1=1\\M_1=0.06mol/dm^3\\V_1=11cm^3=0.011dm^3\\n_2=2\\M_2=?\\V_2=30cm^3=0.030dm^3$ $1cm^3=0.001dm^3$

Putting values in above equation, we get:

$1\times 0.06mol/dm^3\times 0.011dm^3=2\times M_2\times 0.030dm^3\\\\M_2=0.011mol/dm^3$

The concentration in $g/dm^3$ is $0.011mol/dm^3\times 74g/mol=0.814g/dm^3$

Thus concentration of $Ca(OH)_2$ is $0.011mol/dm^3$ and $0.814g/dm^3$

4 0

3 years ago

USE YOUR OWN WORDS. Each question must have a three sentence long answer!

Furkat [3]

Answer:

YOU use YOUR OWN WORDS

Explanation:

Have a good day! :)

8 0

2 years ago

Read 2 more answers

Gallium chloride is formed by the reaction of 2.25 L of a 1.50 M solution of HCl according to the following equation: 2Ga 6HCl -

Setler79 [48]

Answer:

198.56g of GaCl3

Explanation:

We'll begin by calculating the number of mole HCl in 2.25 L of a 1.50 M solution of HCl. This is illustrated below:

Molarity of HCl = 1.50 M

Volume = 2.25 L

Mole of HCl =..?

Molarity = mole /Volume

1.5 = mole /2.25

Cross multiply

Mole = 1.5 x 2.25

Mole of HCl = 3.375 mole

Next, we shall determine the number of mole Gallium chloride, GaCl3 produced from the reaction. This is shown below:

2Ga + 6HCl —> 2GaCl3 + 3H2

From the balanced equation above,

6 moles of HCl reacted to produce 2 moles of GaCl3.

Therefore, 3.375 mole of HCl will react to produce = (3.375 x 2)/6 = 1.125 mole of GaCl3.

Therefore, 1.125 moles of GaCl3 were produced from the reaction.

Next, we shall convert 1.125 mole of GaCl3 to grams. This is illustrated below:

Molar mass of GaCl3 = 70 + (35.5x3) = 176.5g/mol

Mole of GaCl3 = 1.125 mole

Mass of GaCl3 =..?

Mole = mass /Molar mass

1.125 = mass of GaCl3 /176.5

Cross multiply

Mass of GaCl3 = 1.125 x 176.5

Mass of GaCl3 = 198.56g

Therefore, 198.56g of GaCl3 were produced from the reaction.

7 0

3 years ago