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Alenkasestr [34]
4 years ago
15

Consider the following intermediate reactions.

Chemistry
2 answers:
Alja [10]4 years ago
4 0

2.1648 kg of CH4 will generate 119341 KJ of energy.

Explanation:

Write down the values given in the question

CH4(g) +2 O2 → CO2(g) +2 H20 (g)

ΔH1 = - 802 kJ

2 H2O(g)→2 H2O(I)

ΔH2= -88 kJ

The overall chemical reaction is

CH4 (g)+2 O2(g)→CO2(g)+2 H2O (I) ΔH2= -890 kJ

CH4 +2 O2 → CO2 +2 H20

(1mol)+(2mol)→(1mol+2mol)

Methane (CH4) = 16 gm/mol

oxygen (O2) =32 gm/mol

Here 1 mol CH4 ang 2mol of O2 gives 1mol of CO2 and 2 mol of 2 H2O

which generate 882 KJ /mol

Therefore to produce 119341 KJ of energy

119341/882 = 135.3 mol

to produce 119341 KJ of energy, 135.3 mol of CH4 and 270.6 mol of O2 will require

=135.3 *16

=2164.8 gm

=2.1648 kg of CH4

2.1648 kg of CH4 will generate 119341 KJ of energy

schepotkina [342]4 years ago
3 0

Answer:

The answer is A on edge

Explanation:

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7 0
2 years ago
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A balloon with a volume of 2.0 L is filled with a gas at 3 atmospheres. If the pressure is reduced to 0.5 atmospheres without a
creativ13 [48]

Answer:

V2 = 12 L

Explanation:

<u>Given the following data;</u>

Initial volume = 2L

Initial pressure = 3 atm

Final pressure = 0.5 atm

To find the new volume V2, we would use Boyles' law.

Boyles states that when the temperature of an ideal gas is kept constant, the pressure of the gas is inversely proportional to the volume occupied by the gas.

Mathematically, Boyles law is given by;

PV = K

P_{1}V_{1} = P_{2}V_{2}

Substituting into the equation, we have;

3 * 2 = 0.5V_{2}

6 = 0.5V_{2}

V_{2} = \frac {6}{0.5}

V_{2} = 12

<em>V2 = 12L</em>

<em>Therefore, the new volume is 12 Liters. </em>

6 0
3 years ago
The method of changing one or more chemical compounds
horsena [70]

Answer:

chemical process

Explanation:

4 0
3 years ago
How many grams of sodium metal must be introduced to water to produce 3.3 grams of hydrogen gas?
Cloud [144]

Answer:

The mass of sodium metal that must be introduced to water to produce 3.3 grams of hydrogen gas, H₂, is approximately 18.82 grams of sodium metal

Explanation:

The given mass of hydrogen gas produced = 3.3 grams

The molar mass of hydrogen gas, H₂ = 2.016 g/mol

The number of moles of hydrogen gas in 3.3 grams of H₂, 'n', is given as follows;

n = Mass/(Molar mass)

n = 3.3 g/(2.016 g/mol) = 1.63690476 moles of H₂

The reaction of sodium and water can be written as follows;

2Na + 2H₂O → 2NaOH + H₂ (g)

2 moles of sodium produces 1 mole of hydrogen gas, H₂

Therefore;

1.63690476/2 moles of sodium will produce 1.63690476 moles of hydrogen gas, H₂

The molar mass of sodium, Na ≈ 22.989 g/mol

The mass of 1.63690476/2 moles of sodium, 'm', is given as follows;

m = 1.63690476/2 moles × 22.989 g/mol ≈ 18.8154018 grams ≈ 18.82 grams

The mass of sodium that will produce 3.3 grams of hydrogen, m ≈ 18.82 grams of sodium metal.

8 0
3 years ago
Question 8 (9 points) <br><br>Match the lab equipment with its purpose
Deffense [45]

Answer:

\sf \boxed{4} \mapsto Pipet

\sf\boxed{7}\mapsto Test \:tube \: rack

\sf\boxed{3}\mapsto Test\: table

\sf\boxed{5}\mapsto Scoopula

\sf\boxed{1}\mapsto Graduated\: cylinder

\sf\boxed{9}\mapsto Bunsen \:burner

\sf\boxed{2}\mapsto Beaker

\sf \boxed{8}\mapsto Spot\: plate

\sf\boxed{6}\mapsto Goggles

Explanation:

Pipet is used to dispense a very small amount of liquid.

Test tube rack is used to hold multiple test tubes at the same time.

Test Table is used to view chemical reactions or hold or heat small amounts of substance.

Scoopula is used to dispense chemicals from a larger container.

Graduated cylinder is used to measure volume very precisely.

Bunsen burner is used to heat objects.

Beaker is used to transport heat or store substance.

Spot plate is used to observe the color changes of small quantities of a reacting mixture.

Goggles are used to protect the eyes from flying objects or chemical splashes.

_____________________________________________________

7 0
3 years ago
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