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Arisa [49]
3 years ago
15

What is the pH of a 6.2 x 10-5 M HCI solution

Chemistry
1 answer:
Karolina [17]3 years ago
8 0

Answer:

4.21

Explanation:

First, let us generate the dissociation equation. This is illustrated below:

HCl <==> H^+ + Cl^-

From the equation, 1mole of HCl produced 1mole of H^+. This means that 6.2 x 10^-5M will also produce 6.2 x 10^-5M of H^+.

Now we can calculate the pH of the solution as follows:

[H^+] = 6.2 x 10^-5M

pH = —Log [H^+]

PH = —Log 6.2 x 10^-5

pH = 4.21

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Jobisdone [24]

Answer:

m - commerce is the buying and selling of goods and service through wireless handheld devices such as smartphones and tablets as a form of commerce, m - commerce enables users to access online shopping plat forms without needing to use a desktop computer

3 0
3 years ago
The heat of combustion (∆H) for an unknown hydrocarbon is -8.21 kJ/mol. If 0.424 mol of the hydrocarbon is burned in a bomb calo
klio [65]

When 0.424 moles of an unknown hydrocarbon (∆Hc = -8.21 kJ/mol) is burned in a bomb calorimeter (C = 1.12 kJ/°C), the change in the temperature is 3.10 °C.

The heat of combustion (∆Hc) for an unknown hydrocarbon is -8.21 kJ/mol. The heat released by the combustion of 0.424 moles of the hydrocarbon is:

Qc = 0.424 mol \times \frac{(-8.21kJ)}{mol} = -3.48 kJ

According to the law of conservation of energy, the sum of the heat released by the combustion (Qc) and the heat absorbed by the bomb calorimeter (Qb) is zero.

Qc + Qb = 0\\\\Qb = -Qc = 3.48 kJ

Given the heat absorbed by the bomb calorimeter (Qb) and the heat capacity of the bomb calorimeter (C), we can calculate the temperature change (ΔT) using the following expression.

Qb = C \times \Delta T\\\\\Delta T = \frac{Qb}{C} = \frac{3.48 kJ}{1.12 kJ/\° C } = 3.10 \° C

When 0.424 moles of an unknown hydrocarbon (∆Hc = -8.21 kJ/mol) is burned in a bomb calorimeter (C = 1.12 kJ/°C), the change in the temperature is 3.10 °C.

Learn more: brainly.com/question/24245395

8 0
3 years ago
What volume of methane gas at 237 K and 101.33 kPa do you have when the volume is decreased to 0.50 L, with a temperature of 300
Alexxandr [17]

Answer: A volume of 0.592 L of methane gas is required at 237 K and 101.33 kPa when the volume is decreased to 0.50 L, with a temperature of 300 K and a pressure of 151.99 kPa.

Explanation:

Given: T_{1} = 237 K,   P_{1} = 101.33 kPa,      V_{1} = ?

T_{2} = 300 K,      P_{2} = 151.99 kPa,        V_{2} = 0.50 L

Formula used is as follows.

\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}

Substitute the values into above formula as follows.

\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{101.33 kPa \times V_{1}}{237 K} = \frac{151.99 kPa \times 0.50 L}{300 K}\\V_{1} = 0.592 L

Thus, we can conclude that a volume of 0.592 L of methane gas is required at 237 K and 101.33 kPa when the volume is decreased to 0.50 L, with a temperature of 300 K and a pressure of 151.99 kPa.

8 0
3 years ago
Help please it’s due soon just calculate the atomic mass of lithium that’s all i need
Zinaida [17]
6.941u? Sorry if I get it wrong
4 0
3 years ago
Pretend you’re rinding the bus to school in a very cold and snowy day. List 3 different phases changes you might see along the w
Amiraneli [1.4K]

Answer:

Everyone wearing warm clothes.

Everyone with their cheeks red.

Everyone eating or drinking hot food

4 0
2 years ago
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