Please help i will give 20 points

The concentration of hydrogen in soil sample 13 * 10 ^ - 6 * M the soil is tested with a pH meter, what value should the meter r

vovikov84 [41]

Answer:

pH = 4.9

Explanation:

Given data

[H⁺] = 13 × 10⁻⁶ M

The pH is a scale used to determine the acidity or basicity of a solution. The pH is related to the concentration of hydrogen ions through the following expression.

pH = -log [H⁺]

pH = -log 13 × 10⁻⁶

pH = 4.9

Since the pH < 7, the soil is considered to be acid.

6 0

3 years ago

Predict, using Boyle’s Law, what will happen to a balloon that an ocean diver takes to a pressure of 202 kPa (normal atmospheric

kaheart [24]

Explanation:

Balloon that an ocean diver takes to a pressure of 202 k Pa will get reduced in size that is the volume of the balloon will get reduced. This is because pressure and volume of the gas are inversely related to each other.

According to Boyle's law: The pressure of the gas is inversely proportional to the volume occupied by the gas at constant temperature(in Kelvins).

$pressure\propto \frac{1}{volume}$ (At constant temperature)

The pressure beneath the sea is 202 kPa and the atmospheric pressure is 101.3 kPa . This increase in pressure will result in decrease in volume occupied by the gas inside the balloon with decrease in size of a balloon. Hence, the size of the balloon will get reduced at 202 kPa (under sea).

7 0

3 years ago

Calculate the ph of a dilute solution that contains a molar ratio of potassium acetate to acetic acid (pka ???? 4.76) of (a) 2:1

givi [52]

According to Hasselbach-Henderson equation:

$pH=pK_{a}+log\frac{[A^{-}]}{[HA]}$

Here, $[A^{-}]$ is concentration of conjugate base and [HA] is concentration of acid.

In the given problem, conjugate base is $CH_{3}COOK$ and acid is $CH_{3}COOH$ thus, Hasselbach-Henderson equation will be as follows:

$pH=pK_{a}+log\frac{[CH_{3}COOK]}{[CH_{3}COOH]}$ ...... (1)

(a) Ratio of concentration of potassium acetate and acetic acid is 2:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=2$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{2}{1}=5.06$

Therefore, pH of solution is 5.06.

(b) Ratio of concentration of potassium acetate and acetic acid is 1:3 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{3}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{3}=4.28$

Therefore, pH of solution is 4.28.

(c)Ratio of concentration of potassium acetate and acetic acid is 5:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{5}{1}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{5}{1}=5.45$

Therefore, pH of solution is 5.45.

(d) Ratio of concentration of potassium acetate and acetic acid is 1:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=1$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{1}=4.76$

Therefore, pH of solution is 4.76.

(e) Ratio of concentration of potassium acetate and acetic acid is 1:10 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{10}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{10}=3.76$

Therefore, pH of solution is 3.76.

4 0

3 years ago

2 Points What is the oxidation state of Cl in HCIO3? A:+5 B:-1 C:+6 D: -7

Olenka [21]

Answer:

The oxidation state of Cl in HCIO3 is +5

3 0

3 years ago

tert-Butyl alcohol is a solvent with a Kf of 9.10 ∘C/m and a freezing point of 25.5 ∘C. When 0.807 g of an unknown colorless liq

mylen [45]

Answer: ethylene glycol (molar mass = 62.07 g/mol)

Explanation:

Depression in freezing point :

Formula used for lowering in freezing point is,

$\Delta T_f=k_f\times m$

or,

$\Delta T_f=k_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}$

where,

$T_f$ = change in freezing point

$k_f$ = freezing point constant= $9.10^0C/m$

m = molality

Given mass of solute = 0.807 g

Molar mass of solute=? g/mol

weight of solvent in kg= 11.6 g=0.0116 kg

$\Delta T_f=T^{o}_f-T_f=25.5^0C-15.3^0C)=10.2^0C$

$10.2=9.10\times \frac{0.807}{\text{molar mass of solute}\times 0.0116kg}$

${\text{molar mass of solute}}=62.07 g/mol$

Thus the solute is ethylene glycol which has same molecular mass as calculated, i.e 62.07 g/mol.

6 0

3 years ago