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gizmo_the_mogwai [7]
3 years ago
8

A 2.87 g sample of carbon reacts with hydrogen to form 3.41 g of car fuel. What is the empirical formula for the car fuel?

Chemistry
1 answer:
olga_2 [115]3 years ago
5 0

CH₇ is the empirical formula of the car fuel.

Explanation:

To find the empirical formula we use the following algorithm.

First divide each mass the the molar weight of each element:

for carbon 2.87 / 12 = 0.239

for hydrogen 3.41 / 2 = 1.705

And now divide each quantity by the lowest number which is 0.239:

for carbon 0.239 / 0.239 = 1

for hydrogen 1.705 / 0.239 = 7.13 ≈ 7

The empirical formula of the car fuel is CH₇.

I have to tell you that in reality this formula is wrong because is not possible to exist. However the algorithm for finding the empirical formula is right, the problem may reside in the amounts of carbon and hydrogen given.

Learn more about:

empirical formula

brainly.com/question/5297213

#learnwithBrainly

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How many kilojoules of energy would be required to heat a 225g block of aluminum from 23.0 C to 73.5 C?
gulaghasi [49]

Answer:

\boxed {\boxed {\sf 10.2 \ kJ}}

Explanation:

We are asked to find how many kilojoules of energy would be required to heat a block of aluminum.

We will use the following formula to calculate heat energy.

q=mc \Delta T

The mass (m) of the aluminum block is 225 grams and the specific heat (c) is 0.897 Joules per gram degree Celsius. The change in temperature (ΔT) is the difference between the final temperature and the initial temperature.

  • ΔT = final temperature - inital temperature

The aluminum block was heated from 23.0 °C to 73.5 °C.

  • ΔT= 73.5 °C - 23.0 °C = 50.5 °C

Now we know all three variables and can substitute them into the formula.

  • m= 225 g
  • c= 0.897 J/g° C
  • ΔT= 50.5 °C

q= (225 \ g )(0.897 \ J/g \textdegree C)(50.5 \textdegree C)

Multiply the first two numbers. The units of grams cancel.

q= (225 \ g  * 0.897 \ J/g \textdegree C)(50.5 \textdegree C)

q= (225   * 0.897 \ J / \textdegree C)(50.5 \textdegree C)

q= (201.825\ J / \textdegree C)(50.5 \textdegree C)

Multiply again. This time, the units of degrees Celsius cancel.

q= 201.825 \ J * 50.5

q= 10192.1625 \ J

The answer asks for the energy in kilojoules, so we must convert our answer. Remember that 1 kilojoule contains 1000 joules.

\frac { 1  \ kJ}{ 1000 \ J}

Multiply by the answer we found in Joules.

10192.1625 \ J * \frac{ 1 \ kJ}{ 1000 \ J}

10192.1625  * \frac{ 1 \ kJ}{ 1000 }

\frac {10192. 1625}{1000} \ kJ

10.1921625 \ kJ

The original values of mass, temperature, and specific heat all have 3 significant figures, so our answer must have the same. For the number we found, that is the tneths place. The 9 in the hundredth place tells us to round the 1 up to a 2.

10.2 \ kJ

Approximately <u>10.2 kilojoules</u> of energy would be required.

3 0
3 years ago
2CH2(g) + 50 (g) → 400/(g) + 2 H2O(g)
Nadusha1986 [10]

168.96 g of carbon dioxide (CO₂)

Explanation:

The chemical reaction representing the combustion of acetylene:

2 C₂H₂ (g) + 5 O₂ (g)→ 4 CO₂ (g) + 2 H₂O (g)

number of moles = mass / molecular weight

number of moles of acetylene (C₂H₂) = 50 / 26 = 1.92 moles

Taking in account the stoichiometry of the chemical reaction, we devise the following reasoning:

if       2 moles of acetylene (C₂H₂) produces 4 moles of carbon dioxide (CO₂)

then 1.92 moles of acetylene (C₂H₂) produces X moles of carbon dioxide (CO₂)

X = (1.92 × 4) / 2 = 3.84 moles of carbon dioxide (CO₂)

mass = number of moles × molecular weight

mass of carbon dioxide (CO₂) = 3.84 × 44 = 168.96 g

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combustion of hydrocarbons

brainly.com/question/4919676

brainly.com/question/1406903

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6 0
3 years ago
The combustion of propane may be described by the chemical equation C 3 H 8 ( g ) + 5 O 2 ( g ) ⟶ 3 CO 2 ( g ) + 4 H 2 O ( g ) C
Kipish [7]

Answer: 72 grams of O_2(g) are needed to completely burn 19.7 g C_3H_8(g)

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to molecular mass and contains avogadro's number 6.023\times 10^{23} of particles.

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Putting in the values we get:

\text{Number of moles}=\frac{19.7g}{44g/mol}=0.45moles

C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)

According to stoichiometry:

1 mole of C_3H_8 requires 5 moles of oxygen

0.45 moles of C_3H_8 require= \frac{5}{1}\times 0.45=2.25 moles of oxygen

Mass of O_2=moles\times {\text {Molar mass}}=2.25\times 32=72g

72 grams of O_2(g) are needed to completely burn 19.7 g C_3H_8(g)

7 0
3 years ago
Molar mass of a gas lab help?
LuckyWell [14K]

Answer:

The molar mass is determined by applying the Ideal Gas Law, PV = nRT, where P is the pressure (in atm), V is the volume (in L), n is the number of moles of gas, R is the universal gas constant (0.08206 L∙atm/mol∙K), and T is the temperature (in K).

Hope this helps! :)

3 0
4 years ago
Plant and animal cells release energy stored in the bonds of glucose molecules when they perform _________. Question 1 options:
Dmitry_Shevchenko [17]
Cellular respiration
3 0
3 years ago
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