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What happens to the pressure if the gas volume is cut in half while n and I are held constant?

N76 [4]

Answer: You can use Boyle's law, which states that pressure is inversely related to volume when other variables are held constant. If the final pressure of a gas is half of the initial, the volume must double if temperature is to remain the same.

Explanation:

5 0

3 years ago

Which of the following is an example of kinetic energy?

deff fn [24]

Answer:

A. The energy of moving particles near a fire

Explanation:

7 0

3 years ago

An aqueous solution contains 0.23 M potassium hypochlorite.

Arturiano [62]

Answer:

0.22 mol HClO, 0.11mol HBr.

0.25mol NH₄Cl, 0.12 mol HCl

Explanation:

A buffer is defined as a mixture in solution between weak acid and its conjugate base or vice versa.

Potassium hypochlorite (KClO) could be seen as conjugate base of HClO (Weak acid). That means the addition of 0.22 mol HClO will convert the solution in a buffer. HBr reacts with KClO producing HClO, thus, 0.11mol HBr will, also, convert the solution in a buffer. 0.23 mol HBr will react completely with KClO and in the solution you will have only HClO, no a buffering system.

Ammonia (NH₃) is a weak base and its conjugate base is NH₄⁺. That means the addition of 0.25mol NH₄Cl will convert the solution in a buffer. Also, NH₃ reacts with HCl producing NH₄⁺. Thus, addition of 0.12 mol HCl will produce NH₄⁺. 0.25mol HCl consume all NH₃.

5 0

3 years ago

In the laboratory a student combines 26.2 mL of a 0.234 M chromium(III) acetate solution with 10.7 mL of a 0.461 M chromium(III)

Natalka [10]

Answer: The molarity of $Cr^{3+}$ ions in the solution is 0.299 M

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ .....(1)

For chromium (III) acetate:

Molarity of chromium (III) acetate solution = 0.234 M

Volume of solution = 26.2 mL

Putting values in equation 1, we get:

$0.234=\frac{\text{Moles of chromium (III) acetate}\times 1000}{26.2}\\\\\text{Moles of chromium (III) acetate}=\frac{0.234\times 26.2}{1000}=0.00613mol$

1 mole of chromium (III) acetate $(Cr(CH_3COO)_3)$ produces 1 mole of chromium $(Cr^{3+})$ ions and 3 moles of acetate $(CH_3COO^-)$ ions

Moles of $Cr^{3+}\text{ ions}=(1\times 0.00613)=0.00613moles$

For chromium (III) nitrate:

Molarity of chromium (III) nitrate solution = 0.461 M

Volume of solution = 10.7 mL

Putting values in equation 1, we get:

$0.461=\frac{\text{Moles of chromium (III) nitrate}\times 1000}{10.7}\\\\\text{Moles of chromium (III) nitrate}=\frac{0.461\times 10.7}{1000}=0.00493mol$

1 mole of chromium (III) nitrate $(Cr(NO_3)_3)$ produces 1 mole of chromium $(Cr^{3+})$ ions and 3 moles of nitrate $(NO_3^-)$ ions

Moles of $Cr^{3+}\text{ ions}=(1\times 0.00493)=0.00493moles$

For chromium cation:

Total moles of chromium cations = [0.00613 + 0.00493] = 0.01106 moles

Total volume of solution = [26.2 + 10.7] = 36.9 mL

Putting values in equation 1, we get:

$\text{Molarity of }Cr^{3+}\text{ cations}=\frac{0.01106\times 1000}{36.9}\\\\\text{Molarity of }Cr^{3+}\text{ cations}=0.299M/tex]Hence, the molarity of [tex]Cr^{3+}$ ions in the solution is 0.299 M

5 0

3 years ago

Which of those is a balanced chemical equation? (super important)

alexgriva [62]

I believe C is the balanced equation.

7 0

3 years ago

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