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2 years ago

What are two physical characteristics of ionic compounds?

Chemistry

1 answer:

Lorico [155]2 years ago

3 0

Answer:

ionic compounds are solids, possess high melting points, conduct electricity, are hard but brittle and are formed from an ionic bond between one metal and one nonmetal.

Explanation:

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What type of matter is gravel

GREYUIT [131]

Gravel is a solid, opaque, heterogenous mixture.

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2 years ago

How many grams are there in 7.5x10^23 molecules of H2SO4

horsena [70]

Moles of H2SO4= 7.5x10^23/ 6.02x10^23 = 1.25 (3sf) moles of H2SO4

Mass of 1 mole of H2SO4= 98.1g

Therefore mass of 7.5x10^23 molecules of H2SO4= 122.63g

6 0

2 years ago

Name the homologous series for this compound and show its structure ch3ch2cc(ch2)2ch3

kotykmax [81]

Answer:

Alkyne

Explanation:

The homologous series for this compound is the Alkyne because of the presence of the triple bond present between the third and fourth carbon atoms as seen in the picture attached.

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1 year ago

Which is the correct formula for potassium sulfite

rusak2 [61]

Explanation:

K2SO3

hope it helps!

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3 0

2 years ago

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;

(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

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

2 years ago