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gregori [183]
3 years ago
7

What is the [OH-] in a solution that has a [H+] = 4.51 x 10^-6M?

Chemistry
1 answer:
lakkis [162]3 years ago
6 0

Answer:

2.22×10⁻⁹ M

Explanation:

Applying,

[OH⁻][H⁺] = 1×10⁻¹⁴................... Equation 1

Where [OH⁻] = Hydroxyl ion concentration, [H⁺] = Hydrogen ion concentration.

make [OH⁻] the subject of the equation,

[OH⁻] =  1×10⁻¹⁴/[H⁺]............... Equation 2

From the question,

Given: [H⁺] = 4.51×10⁻⁶ M

Substitute into equation 2

[OH⁻] = 1×10⁻¹⁴/4.51×10⁻⁶

[OH⁻] = 2.22×10⁻⁹ M

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A mixture of 0.10 mol of NO, 0.050 mol of H2, and 0.10 mol of H2O is placed in a 1.0- L vessel at 300 K . The following equilibr
umka2103 [35]

Answer:

The concentration of N2 at the equilibrium will be 0.019 M

Explanation:

Step 1: Data given

Number of moles of NO = 0.10 mol

Number of moles of H2 = 0.050 mol

Number of moles of H2O = 0.10 mol

Volume = 1.0 L

Temperature = 300K

At equilibrium [NO]=0.062M

Step 2: The balanced equation

2NO(g) + 2H2(g) → N2(g) + 2H2O(g)

Step 3: Calculate the initial concentration

Concentration = Moles / volume

[NO] = 0.10 mol / 1L = 0.10 M

[H2] = 0.050 mol / 1L = 0.050 M

[H2O] = 0.10 mol / 1L = 0.10 M

[N2] = 0 M

Step 4: Calculate the concentration at the equilibrium

[NO] at the equilibrium is 0.062 M

This means there reacted 0.038 mol (0.038M) of NO

For 2 moles NO we need 2 moles of H2 to produce 1 mol N2 and 2 moles of H2O

This means there will also react 0.038 mol of H2

The concentration at the equilibrium is 0.050 - 0.038 = 0.012 M

There will be porduced 0.038 moles of H2O, this means the final concentration pf H2O at the equilibrium is 0.100 + 0.038 = 0.138 M

There will be produced 0.038/2 = 0.019 moles of N2

The concentration of N2 at the equilibrium will be 0.019 M

5 0
4 years ago
How many grams of oxygen are produced when 7.65 moles of water is decomposed
maksim [4K]

Answer:

The answer to your question is 122.4 g of O₂

Explanation:

Data

mass of O₂ = ?

moles of H₂O = 7.65

Process

1.- Write the balanced chemical reaction

                   2H₂O  ⇒  2H₂  +  O₂

2.- Convert the moles of H₂O to grams

molar mass of H₂O = 2 + 16 = 18 g

                    18 g of H₂O ---------------- 1 mol

                      x                ----------------- 7.65 moles

                      x = (7.65 x 18) / 1

                      x = 137.7 g H₂O

3.- Calculate the grams of O₂

                 36 g of H₂O -------------------- 32 g of O₂

              137.7 g of H₂O -------------------  x

                        x = (32 x 137.7) / 36

                       x = 122.4 g of O₂

 

6 0
4 years ago
What is the most common isotope for element X
ankoles [38]

Answer:

isotope 2

Explanation:

it has the highest percentage abundance

5 0
3 years ago
If you have a gold chain that weighs 80.7 grams, what is the volume of the chain?​
just olya [345]

The volume of the gold chain is 4.18 cm³

<h3>What is density? </h3>

The density of a substance is simply defined as the mass of the subtance per unit volume of the substance. Mathematically, it can be expressed as

Density = mass / volume

With the above formula, we can determine the volume of the chain. Details below:

<h3>How to determine the volume of the gold chain</h3>

The following data were obtained from the question:

  • Mass of gold = 80.7 g
  • Density of gold = 19.3 g/cm³
  • Volume of gold chain =?

Density = mass /volume

Cross multiply

Density × volume = mass

Divide both sides by density

Volume = mass / density

Volume of gold chain = 80.7 / 19.3

Volume of gold chain = 4.18 cm³

Learn more about density:

brainly.com/question/952755

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7 0
2 years ago
3. What is the mass % of a solution that contains 36g KCl in 475g of water?
stira [4]

Answer:

%KCl = 7.05%

%Water = 92.95%

Explanation:

Step 1: Given data

  • Mass of KCl (solute): 36 g
  • Mass of water (solvent): 475 g

Step 2: Calculate the mass of the solution

The mass of the solution is equal to the sum of the masses of the solute and the solvent.

m = 36 g + 475 g = 511 g

Step 3: Calculate the mass percentage of the solution

We will use the following expression.

%Component = mComponent/mSolution × 100%

%KCl = 36 g/511 g × 100% = 7.05%

%Water = 475 g/511 g × 100% = 92.95%

7 0
3 years ago
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