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zheka24 [161]
3 years ago
15

Calculate the amount of CO2 (in kg) released when 1 kg of coal is burned. Assume that carbon content of the coal is 50% by mass.

Chemistry
1 answer:
Delvig [45]3 years ago
4 0

Answer:

1.8321 kg

Explanation:

The given 1 kg of coal contains 50% of the carbon atom by mass. Thus, mass of carbon in coal is \frac {50}{100}\times 1\ kg=0.5\ kg

Also, 1 kg = 1000 g

So, mass of carbon = 500 g

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Molar mass of carbon = 12.0107 g/mol

Moles of methanol = 500 g / 12.0107 g/mol = 41.6295 moles

Considering the reaction:

C+O_2\rightarrow CO_2

From the reaction,

1 mole of C react to form 1 mole of CO_2

So,

41.6295 moles of C react to form 41.6295 moles of CO_2

Moles of CO_2 = 41.6295 moles

Molar mass of CO_2 = 44.01 g/mol

So, Mass = Moles × Molar mass = 41.6295 moles × 44.01 g/mol = 1832.1143 g

Also, 1g = 0.001 kg

<u>So, amount of CO_2 released = 1.8321 kg</u>

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marta [7]

Answer:

0.1

Explanation:

We must first put down the equation of the reaction in order to guide our solution of the question.

2HNO3(aq) + Sr(OH)2(aq) -------> Sr(NO3)2(aq) + 2H2O(l)

Now from the question, the following were given;

Concentration of acid CA= ??????

Concentration of base CB= 0.299M

Volume of acid VA= 17.8ml

Volume of base VB= 24.7ml

Number of moles of acid NA= 2

Number of moles of base NB= 1

From;

CAVA/CBVB= NA/NB

CAVANB= CBVBNA

CA= CBVBNA/VANB

SUBSTITUTING VALUES;

CA= 0.299 × 24.7 ×2 / 17.8×1

CA= 0.8298 M

But;

pH= -log[H^+]

[H^+] = 0.8298 M

pH= -log[0.8298 M]

pH= 0.1

5 0
3 years ago
An aqueous solution is listed as being 33.8% solute by mass with a density of 1.15 g/mL, the molar mass of the solute is 145.6 g
vodomira [7]

Answer:

A) 2.69 M

B) 0.059

Explanation:

A) We have:

33.8% solute by mass= 33.8 g solute/100 g solution

molarity = mol solute/ 1 L solution

molarity= \frac{33.8 g solute}{100 g solution} x \frac{1.15 g solution}{1 ml} x \frac{1 mol solute}{145.6 g solute} x \frac{1000 ml}{1 L}

molarity= 2.69 mol solute/L solution = 2.69 M

B) We know that there are 33.8 g of solute in 100 g of solution.

As the total solution is compounded by solute+solvent (in this case, solvent is water), the mass of water is the difference between the mass of the total solution and the mass of solute:

mass of water= 100 g - 33.8 g = 66.2 g

Now, we calculate the number of mol of both solute and water:

mol solute= 33.8 g solute x \frac{1 mol solute}{145.6 g} = 0.232 mol

mol H20= 66.2 g H₂O x \frac{1 mol H2O}{18 g}

Finally, the mol fraction of solute (Xsolute) is calculated as follows:

Xsolute=\frac{mol solute}{total mol}= \frac{mol solute}{mol solute + mol H2O}=\frac{0.232 mol}{0.232 mol + 3.677 mol}

Xsolute= 0.059

4 0
3 years ago
De selectat din poezia ,,O strada cu sentimente" cat mai multe imagini artistice si figuri de stil
Oliga [24]

Answer:

is this english?

Explanation:

7 0
2 years ago
What mass of ammonia, NH3, is necessary to react with 2.1 x 10^24 molecules of oxygen in the following reaction? 4NH3(g) + 7O2(g
liubo4ka [24]
33.94 
multiply 17.031*1.993
4 0
3 years ago
Does anybody know how to do q4. Please show working out thanks.
insens350 [35]

Answer:

% purity of limestone = 96.53%

Explanation:

Question (4).

Weight of impure CaCO₃ = 25.9 g

Molecular weight of CaCO₃ = 40 + 12 + 3(16)

                                              = 100 g per mole

We know at S.T.P. number of moles of CO₂ = 1 and volume = 22.4 liters

From the given reaction, 1 mole of CaCO₃ reacts with 1 mole or 22.4 liters of

CO₂.

∵ 22.4 liters of CO₂ was produced from CaCO3 = 100 g

∴ 1 liter of CO₂ will be produced by CaCO₃ = \frac{100}{22.4}

∴ 5.6 liters of CO₂ will be produced by CaCO₃ = \frac{100\times 5.6}{22.4}

                                                                              = 25 g

Therefore, % purity of CaCO₃ = \frac{\text{Weight calculated}}{{\text{Weight given}}}\times 100

                                                 = \frac{25}{25.9}\times 100

                                                 = 96.53 %

7 0
3 years ago
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