Care sunt factorii de poluare a aerului?

Calculate [H3O+] and [OH−] for each of the following solutions at 25 ∘C given the pH. pH= 2.89

dimaraw [331]

Answer: The value of $[H_{3}O^{+}]$ is 0.0012 M and $[OH^{-}]$ is $1.02 \times 10^{-14}$ .

Explanation:

pH is the negative logarithm of concentration of hydrogen ion.

It is given that pH is 2.89. So, the value of concentration of hydrogen ions is calculated as follows.

$pH = - log [H^{+}]\\2.89 = - log [H^{+}]\\conc. H^{+} = 0.0012 M$

The relation between pH and pOH value is as follows.

pH + pOH = 14

0.0012 + pOH = 14

pOH = 14 - 0.0012 = 13.99

Now, pOH is the negative logarithm of concentration of hydroxide ions.

Hence, $[OH^{-}]$ is calculated as follows.

$pOH = - log [OH^{-}]\\13.99 = - log [OH^{-}]\\conc. OH^{-} = 1.02 \times 10^{-14} M$

Thus, we can conclude that the value of $[H_{3}O^{+}]$ is 0.0012 M and $[OH^{-}]$ is $1.02 \times 10^{-14}$ .

6 0

2 years ago

An electrolytic cell Group of answer choices consumes electrical current to drive a nonspontaneous chemical reaction. produces e

Rzqust [24]

Answer:

produces electrical current from electricity. produces electrical current from a nonspontaneous chemical reaction.

Explanation:

An electrolytic cell converts electrical energy into chemical energy. Here, the redox reaction is spontaneous and is responsible for the production of electrical energy. The reaction at the anode is oxidation and that at the cathode is reduction. Here, the anode is positive and the cathode is the negative electrode.

3 0

3 years ago

Consider the reaction Mg(s) + I2 (s) → MgI2 (s) Identify the limiting reagent in each of the reaction mixtures below:

Lapatulllka [165]

Answer:

a) Nor Mg, neither I2 is the limiting reactant.

b) I2 is the limiting reactant

c) Mg is the limiting reactant

d) Mg is the limiting reactant

e) Nor Mg, neither I2 is the limiting reactant.

f) I2 is the limiting reactant

g) Nor Mg, neither I2 is the limiting reactant.

h) I2 is the limiting reactant

i) Mg is the limiting reactant

Explanation:

Step 1: The balanced equation:

Mg(s) + I2(s) → MgI2(s)

For 1 mol of Mg we need 1 mol of I2 to produce 1 mol of MgI2

a. 100 atoms of Mg and 100 molecules of I2

We'll have the following equation:

100 Mg(s) + 100 I2(s) → 100MgI2(s)

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

b. 150 atoms of Mg and 100 molecules of I2

We'll have the following equation:

150 Mg(s) + 100 I2(s) → 100 MgI2(s)

I2 is the limiting reactant, and will be completely consumed. There will be consumed 100 Mg atoms. There will remain 50 Mg atoms.

There will be produced 100 MgI2 molecules.

c. 200 atoms of Mg and 300 molecules of I2

We'll have the following equation:

200 Mg(s) + 300 I2(s) →200 MgI2(s)

Mg is the limiting reactant, and will be completely consumed. There will be consumed 200 I2 molecules. There will remain 100 I2 molecules.

There will be produced 200 MgI2 molecules.

d. 0.16 mol Mg and 0.25 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Mg is the limiting reactant, and will be completely consumed. There will be consumed 0.16 mol of I2. There will remain 0.09 mol of I2.

There will be produced 0.16 mol of MgI2.

e. 0.14 mol Mg and 0.14 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

There will be consumed 0.14 mol of Mg and 0.14 mol of I2. there will be produced 0.14 mol of MgI2

f. 0.12 mol Mg and 0.08 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

I2 is the limiting reactant, and will be completely consumed. There will be consumed 0.08 moles of Mg. There will remain 0.04 moles of Mg.

There will be produced 0.08 moles of MgI2.

g. 6.078 g Mg and 63.455 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 6.078 grams / 24.31 g/mol = 0.250 moles

Number of moles I2 = 63.455 grams/ 253.8 g/mol = 0.250 moles

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

There will be consumed 0.250 mol of Mg and 0.250 mol of I2. there will be produced 0.250 mol of MgI2

h. 1.00 g Mg and 2.00 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 1.00 grams / 24.31 g/mol = 0.0411 moles

Number of moles I2 = 2.00 grams/ 253.8 g/mol = 0.00788 moles

I2 is the limiting reactant, and will be completely consumed. There will be consumed 0.00788 moles of Mg. There will remain 0.03322 moles of Mg.

There will be produced 0.00788 moles of MgI2.

i. 1.00 g Mg and 2.00 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 1.00 grams / 24.31 g/mol = 0.0411 moles

Number of moles I2 = 20.00 grams/ 253.8 g/mol = 0.0788 moles

Mg is the limiting reactant, and will be completely consumed. There will be consumed 0.0411 moles of Mg. There will remain 0.0377 moles of I2.

There will be produced 0.0411 moles of MgI2.

4 0

3 years ago

What is the pressure in atm exerted by 2.48 moles of a gas in a 250.0 mL container at 58 degrees celsius

Dvinal [7]

Since the question manages to include moles, pressure, volume, and temperature, then it is evident that in order to find the answer we will have to use the Ideal Gas Equation: PV = nRT (where P = pressure; V = volume; n = number of moles; R = the Universal Constant [0.082 L·atm/mol·K]; and temperature.

First, in order to work out the questions, there is a need to convert the volume to Litres and the temperature to Kelvin based on the equation:
250 mL = 0.250 L
58 °C = 331 K

Also, based on the equation P = nRT ÷ V

⇒ P = (2.48 mol)(0.082 L · atm/mol · K)(331 K) ÷ 0.250 L
⇒ P = (67.31 L · atm) ÷ 0.250 L
⇒ P = 269.25 atm

Thus the pressure exerted by the gas in the container is 269.25 atm.

7 0

2 years ago

What is the density of a sample of a substance with a volume of 120ml and a mass of 90g?

tigry1 [53]

Answer:

<h3>The answer is 0.75 g/mL</h3>

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

From the question

mass = 90 g

volume = 120 mL

We have

$density = \frac{90}{120} = \frac{9}{12} = \frac{3}{4} \\$

We have the final answer as

Hope this helps you

6 0

2 years ago