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3 years ago

Calculate the % Oxygen in NaHCO3

Chemistry

1 answer:

Alexxx [7]3 years ago

3 0

Answer:

Molar mass of NaHCO3 = 84.00661 g/mo

Explanation:

This compound is also known as Baking Soda or Sodium Bicarbonate.

Convert grams NaHCO3 to moles or moles NaHCO3 to grams

Molecular weight calculation:

22.98977 + 1.00794 + 12.0107 + 15.9994*3

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Answer each of the following questions with increases, decreases, or does not change.

Kaylis [27]

Answer:

1) Increases

2) decreases

3) increases

4) decreases

Explanation:

When the intermolecular forces in a liquid increases, the greater vapour pressure of the liquid decreases accordingly.

Since the vapour pressure is proportional to temperature, as temperature increases, the vapour pressure increases alongside.

As intermolecular forces increases, the boiling point increases accordingly since more energy is required to break intermolecular bonds.

Lastly, the greater the surface area, tell greater the vapour pressure since more liquid surface area is now available.

7 0

3 years ago

HELP!!!!!! PLEASE!!!

mixas84 [53]

Answer:

Explanation:

sorry if im wrong!!

7 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

811.68 J of energy are used on a 95g chunk of unknown metal. Calculate the resulting temperature change if the specific heat of

Ivan

Answer:

16 °C

Explanation:

Step 1: Given data

Provided heat (Q): 811.68 J

Mass of the metal (m): 95 g

Specific heat capacity of the metal (c): 0.534 J/g.°C

Step 2: Calculate the temperature change (ΔT) experienced by the metal

We will use the following expression.

Q = c × m × ΔT

ΔT = Q/c × m

ΔT = 811.68 J/(0.534 J/g.°C) × 95 g = 16 °C

6 0

3 years ago

A plant extract can be obtained with organic solvents such as acetone or ethyl alcohol, and is composed of plant pigments such a

prohojiy [21]

Answer:

Here's what I get

Explanation:

A plant extract is a mixture because it contains different substances: acetone or ethanol, chlorophylls A and B, carotene and xanthophylls.

It is homogeneous because it is a solution. There is only one phase: the liquid phase. You cannot see the pigments as separate phases.

You can separate the pigments by paper, thin layer, or column chromatography.

Many schools use paper chromatography, because paper is cheap.

As the mixture of pigments follows the solvent up the paper, they separate into different coloured bands according to their attractive forces to the cellulose in the paper.

The chlorophylls are strongly attracted to the paper, so they don't travel very far.

The nonpolar carotene molecules have little attraction to the polar cellulose, so they are carried along by the solvent front.

8 0

3 years ago