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sergeinik [125]
3 years ago
6

At 1.0atm a gas has a volume of 36.7 L, what is the volume at 5.3atm?

Chemistry
1 answer:
Lapatulllka [165]3 years ago
7 0

Answer:

V_2=6.92L

Explanation:

Hello,

In this case, we use the Boyle's law to understand the volume-pressure relationship as an inversely proportional relationship:

P_1V_1=P_2V_2

In such a way, we compute the resulting final volume as shown below:

V_2=\frac{P_1V_1}{P_2}=\frac{1.0atm*36.7L}{5.3atm} \\\\V_2=6.92L

Best regards.

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How many liters are in a .00813M solution that contains 1.55 g of KBr?
Sloan [31]

Answer:

                      1.602 L (or) 1602 mL

Explanation:

             Molarity is the amount of solute dissolved per unit volume of solution. It is expressed as,

                         Molarity  =  Moles / Volume of Solution    ----- (1)

Rearranging above equation for volume,

                         Volume of solution  =  Moles / Molarity    -------(2)

Data Given;

                  Molarity  =  0.00813 mol.L⁻¹

                  Mass  =  1.55 g

First calculate Moles for given mass as,

                   Moles  =  Mass / M.mass

                   Moles  =  1.55 g / 119.002 g.mol⁻¹

                   Moles  =  0.0130 mol

Now, putting value of Moles and Molarity in eq. 2,

                         Volume of solution  =  0.0130 mol / 0.00813 mol.L⁻¹

                         Volume of solution  = 1.60 L

or,

                         Volume of solution  =  1602 mL

5 0
3 years ago
What is the molarity of a sodium chloride solution made by dissolving 4.512 moles to make 2.0 L?
Svet_ta [14]

Answer:

2.40 M

Explanation:

The molarity of a solution tells you how many moles of solute you get per liter of solution.

Notice that the problem provides you with the volume of the solution expressed in milliliters,

mL

. Right from the start, you should remember that you must convert this volume to liters by using the conversion factor

1 L

=

10

3

mL

Now, in order to get the number of moles of solute, you must use its molar mass. Now, molar masses are listed in grams per mol,

g mol

−

1

, which means that you're going to have to convert the mass of the sample from milligrams to grams

1 g

=

10

3

mg

Sodium chloride,

NaCl

, has a molar mass of

58.44 g mol

−

1

, which means that your sample will contain

unit conversion



280.0

mg

⋅

1

g

10

3

mg

⋅

molar mass



1 mole NaCl

58.44

g

=

0.004791 moles NaCl

This means that the molarity of the solution will be

c

=

n

solute

V

solution

c

=

0.004791 moles

2.00

⋅

10

−

3

L

=

2.40 M

The answer is rounded to three sig figs, the number of sig figs you have for the volume of the solution.

6 0
3 years ago
A sample of carbon dioxide has a pressure of 1.2 atm, a volume of
GalinKa [24]

The answer for the following question is mentioned below.

<u><em>Therefore no of moles present in the gas are 1.12 moles</em></u>

Explanation:

Given:

Pressure of gas (P) = 1.2 atm

Volume of a gas (V) = 50.0 liters

Temperature (T) =650 K

To calculate:

no of moles present in the gas (n)

We know;

According to the ideal gas equation;

We know;

<u>P × V = n × R × T </u>

where,

P represents pressure of the gas

V represents volume of the gas

n represents no of the moles of a gas

R represents the universal gas constant  

where the value of R is 0.0821 L atm  mole^{-1}  K^-1

T represents the temperature of the gas

As we have to calculate the no of moles of the gas;

n = \frac{P*V}{R*T}

n = \frac{1.2*50.0}{0.0821*650}

n = \frac{60}{53.365}

n = 1.12 moles

<u><em>Therefore no of moles present in the gas are 1.12 moles</em></u>

3 0
3 years ago
What must be true of a substance with a lot
LUCKY_DIMON [66]

Answer:

A is the answer.

Explanation:

8 0
3 years ago
Read 2 more answers
What percentage of a radioactive species would be found as daughter material after six half–lives?
Novosadov [1.4K]
100%.....50%.....25%......12.5%......6.25%......3.125%......1.5625%
...........1............2...........3..............4.................5................6..................

After six half-lives would be found 1.5625% of readioactive species.
3 0
3 years ago
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