Write the correct ionic compound formula

How would I complete this table?

defon

Answer:

Fist column, apply the density of water.

Second column is material density, so it's mass of sample / volume of sample

Explanation:

Water density is 1 g/mL

Mass / Volume

If you have 0.25 g, 0.45g and 0.46 g of water, then you have the same values as volume, 0.25mL, 0.45mL and 0.46mL.

Again, density is the relation mass / volume so:

2.20 g / 0.25mL

3.73 g / 0.46 mL

3.52 g / 0.45mL

4 0

3 years ago

In the reaction CuO(s) + CO2(g) → CuCO3(s), a. CO2 is the Lewis acid and CuCO3 is its conjugate base. b. O2– acts as a Lewis bas

adoni [48]

Answer: The correct answer is Option d.

Explanation:

According to Lewis acid-base concept:

The substance which is donating electron pair is considered as Lewis base and the substance which is accepting electron pair is considered as Lewis acid.

For the given chemical reaction:

$CuO(s)+CO_2(g)\rightarrow CuCO_3(s)$

$CO_2$ is accepting electron pair and is getting converted to $CO_3^{2-}$ . Thus, it is considered as Lewis acid.

$O^{2-}$ present in CuO is a Lewis base because it is donating electron pair.

Thus, the correct answer is Option d.

4 0

3 years ago

Which type of wave is this?

mixer [17]

Answer:

A Transverse Wave

8 0

3 years ago

In general, the ________ of a simple machine is the ratio of the distance over which the force is applied to the distance over w

madreJ [45]

The answer that best fits the blank provided above is MECHANICAL.

7 0

3 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;

(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