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3 years ago

The molarity of an aqueous solution of NaCl is defined as the

Chemistry

2 answers:

Andreyy893 years ago

5 0

Answer:

The correct option is (4)

Explanation:

Molarity (M) is generally defined as the number of moles of a solute per liter of the solution.

The formula for molarity (from the definition above) is

M = number of moles (n) ÷ volume (in liters or dm³)

The unit for molarity is mol/L or mol/dm³

From the above, it can be deduced that the molarity of an aqueous solution of common salt (NaCl) is the number of moles of NaCl per liter of the salt solution.

zlopas [31]3 years ago

4 0

Molarity = number of moles / volume in liters solution

Answer (4)

hope this helps!

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A 6.87-L sample of gas has a pressure of 0.732 atm and a temperature of 95 °C. The sample is allowed to expand to a volume of 9.

Jlenok [28]

Answer: The new pressure of the gas, assuming that no gas escaped during the experiment is 0.470 atm

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 0.732 atm

$P_2$ = final pressure of gas = ?

$V_1$ = initial volume of gas = 6.87 L

$V_2$ = final volume of gas = 9.22 L

$T_1$ = initial temperature of gas = $95^oC=273+95=368K$

$T_2$ = final temperature of gas = $44^oC=273+44=317K$

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

$\frac{0.732\times 6.87}{368}=\frac{P_2\times 9.22}{317}$

$P_2=0.470atm$

Thus the new pressure of the gas, assuming that no gas escaped during the experiment is 0.470 atm

7 0

4 years ago

Fill In the blank: Compression is the part of the medium where particles are ________.

STALIN [3.7K]

Compressed is the answer!

I hope this helps. :)

7 0

4 years ago

A gas has a volume of 300 mL in a rigid container at 50oC and 1.75 atm. What will be its pressure at 100K?

eduard

Answer:

Its pressure will be 0.54 atm at 100 K.

Explanation:

Gay-Lussac's law indicates that, as long as the volume of the container containing the gas is constant, as the temperature increases, the gas molecules move faster. Then the number of collisions with the walls increases, that is, the pressure increases. That is, the pressure of the gas is directly proportional to its temperature.

Gay-Lussac's law can be expressed mathematically as the quotient between pressure and temperature equal to a constant:

$\frac{P}{T} =k$

Studying two different states, an initial state 1 and a final state 2, it is satisfied:

$\frac{P1}{T1} =\frac{P2}{T2}$

In this case:

P1= 1.75 atm
T1= 50 °C= 323 K (being 0 C=273 K)
P2= ?
T2= 100 K

Replacing:

$\frac{1.75 atm}{323 K} =\frac{P2}{100 K}$

Solving:

$P2= 100 k*\frac{1.75 atm}{323 K}$

P2= 0.54 atm

Its pressure will be 0.54 atm at 100 K.

8 0

3 years ago

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4 0

3 years ago

Write a balanced equation for the complete oxidation reaction that occurs when methane (CH4) burns in air.

Lelechka [254]

Answer: The balanced equation for the complete oxidation reaction that occurs when methane (CH4) burns in air is $CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O$ .

Explanation:

When a substance tends to gain oxygen atom in a chemical reaction and loses hydrogen atom then it is called oxidation reaction.

For example, chemical equation for oxidation of methane is as follows.

$CH_{4} + O_{2} \rightarrow CO_{2} + H_{2}O$

Number of atoms present on reactant side are as follows.

C = 1
H = 4
O = 2

Number of atoms present on product side are as follows.

C = 1
H = 2
O = 3

To balance this equation, multiply $O_{2}$ by 2 on reactant side. Also, multiply $H_{2}O$ by 2 on product side. Hence, the equation can be rewritten as follows.

$CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O$

Now, the number of atoms present on reactant side are as follows.

C = 1
H = 4
O = 4

Number of atoms present on product side are as follows.

C = 1
H = 4
O = 4

Since, the atoms present on both reactant and product side are equal. Therefore, this equation is now balanced.

Thus, we can conclude that balanced equation for the complete oxidation reaction that occurs when methane (CH4) burns in air is $CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O$ .

4 0

3 years ago