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3 years ago

Why must all chemical equations be balanced?

1 answer:

8 0

If the equation is not balanced, you have extra of one substance. If you try to do an experiment, you could wind up with a totally different outcome. Its kind of like baking a cake. If the ingredients aren't the right amounts, you dont get an edible cake.

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The reaction for the formation of gaseous hydrogen fluoride (HF) from molecular hydrogen (H2) and fluorine (F2) has an equilibri

mestny [16]

Answer:

[H2]eq = 0.0129 M

[F2]eq = 1.0129 M

[HF]eq = 0.9871 M

Explanation:

H2(g) + F2(g) ↔ 2HF(g)

∴ Ke = [HF]² / [H2]*[F2] = 1.15 E2

experiment:

∴ n H2 = 3.00 mol

∴ n F2 = 6.00 mol

∴ V sln = 3.00 L

⇒ [H2]i = 3.00 mol / 3.00 L = 1 M

⇒ [F2]i = 6.00 mol / 3.00 L = 2 M

[ ]i change [ ]eq

H2 1 1 - x 1 - x

F2 2 2 - x 2 - x

HF - x x

⇒ K = (x)² / (1 - x)*(2 - x) = 1.15 E2

⇒ x² / (2 - 3x + x²) = 1.15 E2 = 115

⇒ x² = (2 - 3x + x²)(115)

⇒ x² = 230 - 345x + 115x²

⇒ 0 = 230 - 345x + 114x²

⇒ x = 0.9871

equilibrium:

⇒ [H2] = 1 - x = 1 - 0.9871 = 0.0129 M

⇒ [F2] = 2 - x = 2 - 0.9871 = 1.0129 M

⇒ [HF] = x = 0.9871 M

5 0

3 years ago

Box 1

Alex73 [517]

10g

Explanation:

Box 1, Mass of A = 10g

Box 2, Mass of B = 5g

Box 3, = 1A + 1B

Unknown:

Mass of B that would combine with mass of 20g of A

Solution:

Mass ratio of A to B:

$\frac{mass of A}{mass of B}$ = mass ratio

$\frac{10}{5}$ = mass ratio

The mass ratio of A to B = 2: 1

Now, number of B that will combine with 20g of A;

$\frac{mass of A}{mass of B}$ = mass ratio

$\frac{20}{mass of B}$ = $\frac{2}{1}$

Mass of B = 10g

10g of B would combine with 20g of A

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7 0

3 years ago

Read 2 more answers

What are the primary elements of the earths layres?

Vaselesa [24]

The 4 main one are Iron, oxygen, silicon, and magnesium

4 0

3 years ago

a sample of helium occupies a volume of 101.2 mL at a pressure of 790 mmHg. at what pressure would the volume be 120 mL?

AveGali [126]

Answer : The final pressure will be, 666.2 mmHg

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

$P\propto \frac{1}{V}$

or,

$P_1V_1=P_2V_2$

where,

$P_1$ = initial pressure = 790 mmHg

$P_2$ = final pressure = ?

$V_1$ = initial volume = 101.2 mL

$V_2$ = final volume = 120 mL

Now put all the given values in the above equation, we get:

$790mmHg\times 101.2mL=P_2\times 120mL$

$P_2=666.2mmHg$

Therefore, the final pressure will be, 666.2 mmHg

6 0

3 years ago

When carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 14.4 g of carbon were burned in the presence of

Natasha2012 [34]

When carbon reacts with oxygen it forms CO2. This can depicted by the below equation.

C + O2→ CO2

It has been mentioned that when 14.4 g of C reacts with 53.9 g of O2, then 15.5 g of O2 remains unreacted. This indicates that Carbon is the limiting reagent and hence the amount of CO2 produced is based on the amount of Carbon burnt.

C + O2→ CO2

In the above equation , 1 mole of carbon reacts with 1 mole of O2 to produce 1 mole of CO2.

In this case 14.4 g of Carbon reacts with 53.9 of O2 to produce "x"g of CO2.

No of moles = mass of the substance÷molar mass of the substance

No of moles of carbon = 14.4 /12= 1.2 moles

No of moles of O2 = Mass of reacted O2/Molar mass of O2.

No of moles of O2 = (Total mass of O2 burned - Mass of unreacted O2)/32

No of moles of O2 = (53.9-15.5) ÷ 32 = 1.2 moles.

Hence as already discussed 1 mole of Carbon reacts with 1 mole of O2 to produce 1 mole of CO2. In this case 1.2 moles of carbon reacts with 1.2 moles of O2 to produce 1.2 moles of CO2.

Moles of carbon dioxide = Mass of CO2 produced /Molar mass of CO2

Mass of CO2 produced(x) = Moles of CO2 ×Molar mass of CO2

Mass of CO2 produced(x) = 1.2 x 44 = 52.8 g

Thus 52.8 g of CO2 is produced.

5 0

3 years ago