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3 years ago

What is the molarity of a solution that contains 0.220 moles KOH in 0.350 L of solution?

Chemistry

1 answer:

Lera25 [3.4K]3 years ago

5 0

Answer:

0.628 M.

Explanation:

In order to solve this problem we need to keep in mind the definition of molarity:

Molarity = moles / liters

We are given both the number of moles and the volume of solution, meaning we can now proceed to calculate the molarity:

Molarity = 0.220 mol / 0.350 L

Molarity = 0.628 M

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Which one of the following statements is true?

sergij07 [2.7K]

Answer:

A calculator has an endifiite shape because all of its atoms are touching each other

8 0

3 years ago

A chemist measures the energy change ΔH during the following reaction: C3H8 (g) +5O2 (g) →3CO2 (g) +4H2O (l) =ΔH−2220.kJ Use the

statuscvo [17]

Answer:

The reaction is exothermic.

Yes, released.

The heat released is 4,08x10³ kJ.

Explanation:

For the reaction:

C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l)

The ΔH is -2220 kJ, As ΔH is <0, The reaction is exothermic.

As the reaction is exothermic, the heat of the reaction will be released.

The heat released in 81,0g is:

81,0g C₃H₈× $\frac{1mol}{44,1g}$ × $\frac{2220kJ}{1mol}$ = 4,08x10³ kJ

-Using molar mass of C₃H₈ to convert mass to moles and knowing that there are released 2220 kJ per mole of C₃H₈-

I hope it helps!

3 0

3 years ago

I have to do this for homework please help :)

kramer

Answer:

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6 0

3 years ago

The molarity (M) of an aqueous solution containing 22.5 g of sucrose (C12H22O11) in 35.5 mL of solution is ________.

Nonamiya [84]

Answer:

1.86 M

Explanation:

From the question given above, the following data were obtained:

Mass of sucrose (C12H22O11) = 22.5 g

Volume of solution = 35.5 mL

Molarity of solution =?

Next, we shall determine the number of mole in 22.5 g of sucrose (C12H22O11). This can be obtained as follow:

Mass of sucrose (C12H22O11) = 22.5 g

Molar mass of C12H22O11 = (12×12) + (22×1) + (16×11)

= 144 + 22 + 176

= 342 g/mol

Mole of C12H22O11 =?

Mole = mass /Molar mass

Mole of C12H22O11 = 22.5 /342

Mole of sucrose (C12H22O11) = 0.066 mole

Next, we shall convert 35.5 mL to litres (L). This can be obtained as follow:

1000 mL = 1 L

Therefore,

35.5 mL = 35.5 mL × 1 L / 1000 mL

35.5 mL = 0.0355 L

Thus, 35.5 mL is equivalent to 0.0355 L.

Finally, we shall determine the molarity of the solution as follow:

Mole of sucrose (C12H22O11) = 0.066 mole

Volume of solution = 0.0355 L.

Molarity of solution =?

Molarity = mole /Volume

Molarity of solution = 0.066/0.0355

Molarity of solution = 1.86 M

Therefore, the molarity of the solution is 1.86 M.

8 0

3 years ago

Which law relates to the ideal gas law?

professor190 [17]

Answer: The law that related the ideal gas law is $\frac{V_1}{n_1}=\frac{V_2}{n_2}$

Explanation:

There are 4 laws of gases:

Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature.

Mathematically,

$P_1V_1=P_2V_2$

Charles' Law: This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

Gay-Lussac Law: This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

Avogadro's Law: This law states that volume is directly proportional to number of moles at constant temperature and pressure.

Mathematically,

$\frac{V_1}{n_1}=\frac{V_2}{n_2}$

Hence, the law that related the ideal gas law is $\frac{V_1}{n_1}=\frac{V_2}{n_2}$

8 0

3 years ago

Read 2 more answers