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3 years ago

Glycerol (molar mass 92.09 g/mol) has been suggested for use as an alternative fuel. The enthalpy of combustion, Δ

1 answer:

6 0

The mass of glycerol to that would need to be combusted to heat 500.0g of water from 20.0°C to 100.0°C is; 9.32 grams.

We must establish the fact that energy is neither created nor destroyed.

Therefore, the amount of heat absorbed by water is equal to the amount of heat released by the combustion of glycerol.

Total heat absorbed by water, H(water) is;

Q(water) = m C (T2 - T1)

Q(water) = 500 × 4.184 × (100-20)

Q(water) = 167.36 kJ

Consequently, the quantity of heat evolved by the combustion of glycerol is;

Q(glycerol) = 167,360 J = n × ΔH°comb

where, n = no. of moles of glycerol.

167.36 kJ= n × 1654 kJ/mole

n = 167.36/1654

n = 0.1012 moles of glycerol.

Therefore, mass of glycerol combusted, m is;

m = n × Molar mass

m = 0.1012 × 92.09

m = 9.32 g.

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In Haber’s process, 30 moles of hydrogen and 30 moles of nitrogen react to make ammonia. If the yield of the product is 50%, wha

devlian [24]

Answer:

Therewill be produced 170.6 grams NH3, there will remain 25 moles of N2, this is 700 grams

Explanation:

Step 1: Data given

Number of moles hydrogen = 30 moles

Number of moles nitrogen = 30 moles

Yield = 50 %

Molar mass of N2 = 28 g/mol

Molar mass of H2 = 2.02 g/mol

Molar mass of NH3 = 17.03 g/mol

Step 2: The balanced equation

N2 + 3H2 → 2NH3

Step 3: Calculate limiting reactant

For 1 mol of N2, we need 3 moles of H2 to produce 2 moles of NH3

Hydrogen is the limiting reactant.

The 30 moles will be completely be consumed.

N2 is in excess. There will react 30/3 =10 moles

There will remain 30 -10 = 20 moles (this in the case of a 100% yield)

In a 50 % yield, there will remain 20 + 0,5*10 = 25 moles. there will react 5 moles.

Step 4: Calculate moles of NH3

There will be produced, 30/ (3/2) = 20 moles of NH3 (In case of 100% yield)

For a 50% yield there will be produced, 10 moles of NH3

Step 5: Calculate the mass of NH3

Mass of NH3 = mol NH3 * Molar mass NH3

Mass of NH3 = 20 moles * 17.03

Mass of NH3 = 340.6 grams = theoretical yield ( 100% yield)

Step 6: Calculate actual mass

50% yield = actual mass / theoretical mass

actual mass = 0.5 * 340.6

actual mass = 170.3 grams

Step 7: The mass of nitrogen remaining

There remain 20 moles of nitrogen + 50% of 10 moles = 25 moles remain

Mass of nitrogen = 25 moles * 28 g/mol

Mass of nitrogen = 700 grams

6 0

3 years ago

Read 2 more answers

HCHO2 + NaOH = NaCHO2 + H2O complete ionic

Llana [10]

Answer:

HCHO₂(aq) + Na⁺(aq) + OH⁻(aq) ⟶ Na⁺(aq) + CHO₂⁻(aq) + H₂O(ℓ)

Explanation:

An ionic equation uses the symbols (aq) [aqueous] to indicate molecules and ions that are soluble in water, (s) [solid] to indicate insoluble solids, and (ℓ) to indicate substances (usually water) in the liquid state.

In this reaction, aqueous sodium hydroxide reacts with aqueous formic acid to form sodium formate and water.

1. Molecular equation

HCHO₂(aq) + NaOH(aq) ⟶ NaCHO₂(aq) + H₂O(ℓ)

2. Ionic equation

Sodium hydroxide and sodium formate are soluble ionic compounds, so we write them as hydrated ions.

HCHO₂(aq) + Na⁺(aq) + OH⁻(aq) ⟶ Na⁺(aq) + CHO₂⁻(aq) + H₂O(ℓ)

4 0

3 years ago

Radon-219 decays to radon-218 by releasing... a. a positron b.a muon O c. a neutron O d. an electron O e. a proton

MrRa [10]

Answer: Option (c) is the correct answer.

Explanation:

A positron is denoted as $^{0}_{+1}e$ , a neutron is denoted as $^{1}_{0}n$ , an electron is denoted as $^{0}_{-1}e$ , and a proton is denoted as $^{1}_{1}p$ .

Therefore, when $^{219}Rn$ will decay into $^{218}Rn$ this means there occurs decrease in its mass number. Hence, it means a neutron will be releasing during this reaction.