Could be With this tool, the separated from the

Brainliest for correct answer please show all work

Korvikt [17]

Answer:

1) Na₃PO₄ + 3 KOH ➙ 3 NaOH + K₃PO₄

2) MgF₂ + Li₂CO₃➙ MgCO₃ + 2 LiF

3) P₄ + 3 O₂➙ 2 P₂O₃

Explanation:

To balance an equation, ensure that the number of atoms of each element is equal on both sides.

Reactants would be those on the left of the arrow while products are on the right of the arrow.

Balance O and H atoms last.

Question 1:

__Na₃PO₄ + __KOH ➙ __NaOH + __K₃PO₄

Reactants: 3Na, 1P, 1K, 5O, 1H

Products: 1Na, 1P, 3K, 5O, 1H

Balance the number of Na:

__Na₃PO₄ + __KOH ➙ 3 NaOH + __K₃PO₄

Reactants: 3Na, 1P, 1K, 5O, 1H

Products: 3Na, 1P, 3K, 7O, 3H

Balance the number of K:

__Na₃PO₄ + 3 KOH ➙ 3 NaOH + __K₃PO₄

Reactants: 3Na, 1P, 3K, 7O, 3H

Products: 3Na, 1P, 3K, 7O, 3H

The equation is now balanced.

Question 2:

__MgF₂ + __Li₂CO₃➙ __MgCO₃ + __LiF

Reactants: 1Mg, 2F, 2Li, 1C, 3O

Products: 1Mg, 1F, 1Li, 1C, 3O

Balance number of Li and F atoms:

__MgF₂ + __Li₂CO₃➙ __MgCO₃ + 2 LiF

Reactants: 1Mg, 2F, 2Li, 1C, 3O

Products: 1Mg, 2F, 2Li, 1C, 3O

The equation is now balanced.

Question 3:

__P₄ + __O₂➙ __P₂O₃

Reactants: 4P, 2O

Products: 2P, 3O

Balance the number of P atoms:

__P₄ + __O₂➙ 2 P₂O₃

Reactants: 4P, 2O

Products: 4P, 6O

Balance the number of O atoms:

__P₄ + 3 O₂➙ 2 P₂O₃

Reactants: 4P, 6O

Products: 4P, 6O

The equation is now balanced.

4 0

3 years ago

Combustion analysis is performed on 0.50 g of a hydrocarbon, and 1.47 g of CO2 and 0.902 g of H2O are produced. What is the empi

JulijaS [17]

1. The empirical formula of the hydrocarbon is CH₃

2. The molecular formula of the hydrocarbon is C₂H₆

<h3>How to determine the mass of Carbon </h3>

Mass of CO₂ = 1.47 g
Molar mass of CO₂ = 44 g/mol
Molar of C = 12 g/mol
Mass of C =?

Mass of C = (12 / 44) × 1.47

Mass of C = 0.4 g

<h3>How to determine the mass of H</h3>

Mass of compound = 0.5 g
Mass of C = 0.4 g
Mass of H = ?

Mass of H = (mass of compound) – (mass of C)

Mass of H = 0.5 – 0.4

Mass of H =0.1 g

<h3>1. How to determine the empirical formula </h3>

C = 0.4 g
H = 0.1 g
Empirical formula =?

Divide by their molar mass

C = 0.4 / 12 = 0.03

H = 0.1 / 1 = 0.1

Divide by the smallest

C = 0.03 / 0.03 = 1

H = 0.1 / 0.03 = 3

Thus, the empirical formula of the compound is CH₃

<h3>2. How to determine the molecular formula</h3>

Empirical formula = CH₃
Molar mass = 30 g/mol
Molecular formula =?

Molecular formula = empirical × n = mass number

[CH₃]n = 30

[12 + (3×1)]n = 30

15n = 30

Divide both side by 15

n = 30 / 15

n = 2

Molecular formula = [CH₃]n

Molecular formula = [CH₃]₂

Molecular formula = C₂H₆

Learn more about empirical formula:

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4 0

2 years ago

NH4NO3 (8) N20 (9) + H20 (1)

Ludmilka [50]

Answer:

NH4NO3 = 1

N2O = 1

H2O = 2

Explanation:

3 0

3 years ago

Which statements describe density? Check all that apply.

MrRa [10]

I think is 5 because you use density is found by mass divide by volume.

8 0

3 years ago

Read 2 more answers

Please work out all the problems on this page, no links, no i don't knows please do not take advantage of me or my points please

likoan [24]

Answer:

first one-liquids dont have a fixed shape and solids do

6 0

3 years ago

Read 2 more answers