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2 years ago

Calculate the pH when (a) 24.9 mL and (b) 25.1 mL of 0.100 M HNO3 have been added to 25.0 mL of 0.100 M KOH solution.

Chemistry

1 answer:

kogti [31]2 years ago

7 0

Answer:

Suppose you added some solid NaCl to a saturated solution of NaCl at 20℃ and warmed the mixture to 40℃. What would happen to the added NaCl?

Explanation:

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A sample of nitrogen gas, (N2), occupies 45.0 mL at 27.00∘C and 80.0 kPa. What will be the pressure if the gas is cooled to −73.

maxonik [38]

Answer:

The new pressure is 53.3 kPa

Explanation:

This problem can be solved by this law. when the volume remains constant, pressure changes directly proportional as the Aboslute T° is modified.

T° increase → Pressure increase

T° decrease → Pressure decrease

In this case, temperature was really decreased. So the pressure must be lower.

P₁ / T₁ = P₂ / T₂

80 kPa / 300K = P₂/200K

(80 kPa / 300K) . 200 K = P₂ → 53.3 kPa

6 0

2 years ago

The lab procedure mentions that not all of the oxone may dissolve after being added to the water because the water may be satura

ollegr [7]

Answer:

The answer is (C) There are more solute molecules than water molecules.

Explanation:

A saturated solution is one in which no more solute can be dissolved or disintegrated into the solvent. When or if the ozone stops being dissolved in the water, it implies that the water has already taken on more ozone molecules than it can contain, meaning there are more solute molecules (ozone molecules) than there are solvent molecules (water molecules).

3 0

2 years ago

Read 2 more answers

A student placed 10.5 g of glucose (C6H12O6) in a volumetric fla. heggsk, added enough water to dissolve the glucose by swirling

aniked [119]

Answer: The mass of glucose in final solution is 0.420 grams

Explanation:

To calculate the molarity of solution, we use the equation:

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

Initial mass of glucose = 10.5 g

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$\text{Initial molarity of glucose}=\frac{10.5\times 1000}{180.16\times 100}\\\\\text{Initial molarity of glucose}=0.583M$

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated glucose solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted glucose solution

We are given:

$M_1=0.583M\\V_1=20.0mL\\M_2=?M\\V_2=0.5L=500mL$

Putting values in above equation, we get:

$0.583\times 20=M_2\times 500\\\\M_2=\frac{0.583\times 20}{500}=0.0233M$

Now, calculating the mass of final glucose solution by using equation 1:

Final molarity of glucose solution = 0.0233 M

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$0.0233=\frac{\text{Mass of glucose in final solution}\times 1000}{180.16\times 100}\\\\\text{Mass of glucose in final solution}=\frac{0.0233\times 180.16\times 100}{1000}=0.420g$