Compared to pure water, an aqueous solution of calcium chloride has what?

Formula for water what does the lack of a subscript after those O indicate

photoshop1234 [79]

It indicates that there is only one oxygen molecule

3 0

3 years ago

What is the primary requirement for a molecule to be analyzed by Gas Chromatography?

BabaBlast [244]

Answer:

b) The molecule has a molecular weight under 200 g/mole

Explanation:

The molecule has a molecular weight under 200 g/mole is the primary requirement for a molecule to be analyzed by Gas Chromatography.

3 0

3 years ago

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A power plant is driven by the combustion of a complex fossil fuel having the formula C11H7S. Assume the air supply is composed

AlekseyPX

(a) 4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 20.68N_2;

(b) 4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2;

(c) 23 900 kg air; (d) air:fuel = 10.2; (e) air:fuel = 12.2:1

(a) Balanced equation including N_2 from air

The balanced equation ignoring N_2 from air is

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2

Moles of N_2 =55 mol O_2 × (3.76 mol N_2/1 mol O_2) = 206.8 mol N_2

Including N_2 from air, the balanced equation is

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 206.8N_2

(b) Balanced equation for 120 % stoichiometric combustion

Moles of O_2 = 55 mol O_2 × 1.20 = 66.00 mol O_2

Excess moles O_2 = (66.00 – 55) mol O_2 = 11.00 mol O_2

Moles of N_2 = 66.00 mol O_2 × (3.76 mol N_2/1 mol O_2) = 248.2 mol N_2

The balanced equation is

4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2

(c) Minimum mass of air

Moles of O_2 required = 1700 kg C_11 H_7S

× (1 kmol C_11 H_7S/185.24 kg C_11 H_7S) × (55 kmol O_2/4 kmol C_11 H_7S)

= 126.2 kmol O_2

Mass of O_2 = 126.2 kmol O_2 × (32.00 kg O_2/1 kmol O_2) = 4038 kg O_2

Mass of N_2 required = 126.2 kmol O_2 × (3.76 kmol N_2/1 kmol O_2)

× (28.01 kg N_2/1 kmol N_2) = 13 285 kg N_2

Mass of air = Mass of N_2 + mass of O_2 = (4038 + 13 285) kg = 17 300 kg air

(d) Air:fuel mass ratio for 100 % combustion

Air:fuel = 17 300 kg/1700 kg = 10.2 :1

(e) Air:fuel mass ratio for 120 % combustion

Mass of air = 17 300 kg × 1.20 = 20 760 kg air

Air:fuel = 20 760 kg/1700 kg = 12.2 :1

6 0

3 years ago

Which notation is used to represent a beta particle?

Lapatulllka [165]

The beta particle is represented by the Greek letter β. The beta particle is one of the particles that are emitted during radioactive decay. Beta particles are emitted in the process of beta decay. A beta particle can be in form of an electron or a positron.

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3 years ago

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How much pv work is done in kilojoules for the reaction of 0.68 mol of h2 with 0.34 mol of o2 at atmospheric pressure if the vol

dsp73

The delta H of -484 kJ is the heat given off when 2 moles of H2 react with 1 mole of O2 to make 2 moles of H2O. You don't have anywhere near that much reactants, only 1/4 as much

actual delta H = 0.34 moles H2 x (-484 kJ / 2 moles H2) = 823 kJ 

delta E = delta H - PdeltaV = 823 kJ - 0.41 kJ = 822 kJ

3 0

3 years ago