All categories

2 years ago

5. The compound is called:

Chemistry

1 answer:

motikmotik2 years ago

4 0

Answer:

Vanadium (VI) Phosphide

Explanation:

Vanadium has a 6+ charge for this compound, so there needs to be two phosphorus atoms to satisfy the charges. This is an ionic compound as well so the roman numerals need to be present.

You might be interested in

The theory used to explain the behavior of solids, liquids and gases is

Annette [7]

Kinetic molecular theory.

5 0

3 years ago

Pentane is a small liquid hydrocarbon, between propane and gasoline in size at C5H12. The density of pentane is 0.626 g/mL and i

Goshia [24]

Answer : The fuel value and the fuel density of pentane is, 49.09 kJ/g and $3.073\times 10^4kJ/L$ respectively.

Explanation :

Fuel value : It is defined as the amount of energy released from the combustion of hydrocarbon fuels. The fuel value always in positive and in kilojoule per gram (kJ/g).

As we are given that:

$\Delta H^o_{comb}=-3535kJ/mol$

Fuel value = $\frac{\Delta H^o_{comb}}{\text{Molar mass of pentane}}$

Molar mass of pentane = 72 g/mol

Fuel value = $\frac{3535kJ/mol}{72g/mol}$

Fuel value = 49.09 kJ/g

Now we have to calculate the fuel density of pentane.

Fuel density = Fuel value × Density

Fuel density = (49.09 kJ/g) × (0.626g/mL)

Fuel density = 30.73 kJ/mL = $3.073\times 10^4kJ/L$

Thus, the fuel density of pentane is $3.073\times 10^4kJ/L$

4 0

3 years ago

No FILES PLZZ

Bumek [7]

Answer:

Converted to renewable ones

exhausted or depleted

5 0

2 years ago

Read 2 more answers

Oxygen gas can be prepared by heating potassium chlorate according to the following equation:2KClO3(s)Arrow.gif2KCl(s) + 3O2(g)T

Eduardwww [97]

Answer:

Moles of potassium chlorate reacted = 0.2529 moles

The amount of oxygen gas collected will be 12.8675 g

Explanation:

(a)

We are given:

Vapor pressure of water = 17.5 mmHg

Total vapor pressure = 748 mmHg

Vapor pressure of Oxygen gas = Total vapor pressure - Vapor pressure of water = (748 - 17.5) mmHg = 730.5 mmHg

To calculate the amount of Oxygen gas collected, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 730.5 mmHg

V = Volume of the gas = 9.49 L

T = Temperature of the gas = $20^oC=[20+273]K=293K$

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

$730.5mmHg\times 9.49L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 293K\\\\n=\frac{730.5\times 9.49}{62.3637\times 293}=0.3794mol$

According to the reaction shown below as:-

$2KClO_3(s)\rightarrow 2KCl(s) +3O_2(g)$

3 moles of oxygen gas are produced when 2 moles of potassium chlorate undergoes reaction.

So,

0.3794 mol of oxygen gas are produced when $\frac{2}{3}\times 0.3794$ moles of potassium chlorate undergoes reaction.

Moles of potassium chlorate reacted = 0.2529 moles

(b)

We are given:

Vapor pressure of water = 17.5 mmHg

Total vapor pressure = 753 mmHg

Vapor pressure of Oxygen gas = Total vapor pressure - Vapor pressure of water = (753 - 17.5) mmHg = 735.5 mmHg

To calculate the amount of Oxygen gas collected, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 735.5 mmHg

V = Volume of the gas = 9.99 L

T = Temperature of the gas = $20^oC=[20+273]K=293K$

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

$735.5mmHg\times 9.99L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 293K\\\\n=\frac{735.5\times 9.99}{62.3637\times 293}=0.40211mol$

Moles of Oxygen gas = 0.40211 moles

Molar mass of Oxygen gas = 32 g/mol

Putting values in above equation, we get:

$0.037mol=\frac{\text{Mass of Oxygen gas}}{2g/mol}\\\\\text{Mass of Oxygen gas}=(0.40211mol\times 32g/mol)=12.8675g$

Hence, the amount of oxygen gas collected will be 12.8675 g

5 0

2 years ago

Which one is you fav one

serious [3.7K]

Answer:

the 3rd one

Explanation:

8 0

2 years ago

Read 2 more answers