How many moles of PCl3 are produced if 5 moles of P4 reacted with 22 moles of Cl2

For a closed system, the change in entropy for a reversible process is _________ the change in entropy for an irreversible proce

Grace [21]

Answer:

c. the same as

Explanation:

for a closed system:

⇒ ΔS > 0.....irreversible process

⇒ ΔSrev = ∫ dQ/T

for an irreversible process, ΔS is not the same as in a reversible process, since more than one reversible process is needed for the two ΔS to be equal.

⇒ ΔSirrev = ΔSrev1 + ΔSrev2 + ....

but if the initial and final states are the same, ΔS for an irreversible process can be calculated as if it were a reversible process.

⇒ ΔSirrev = ΔSrev

5 0

3 years ago

. A mixture of methane and air is capable of being ignited only if the mole percent of methane is between 5% and 15%. A mixture

postnew [5]

Answer:

A) Mass flow rate of air = 22.892 kmol/hr

B)percentage by mass of oxygen in the product gas = 22.52%

Explanation:

We are given that the mixture containing 9.0 mole% methane in air flowing.

Thus, we have 0.09 mole of methane(CH4) and the remaining will be the air which is (100% - 9%) = 91% = 0.91

Molar mass of CH4 = 12 + 1(4) = 16 g/mol

We are given the average molecular weight of air = 29 g/mol

Thus;

Average molar mass of air and methane mixture is;

M_avg = (0.09 × 16) + (0.91 × 29)

M_avg = 27.83 g/mol

We are told that air flowing at a rate of 7 × 10² kg/h = 700 kg/h

Thus;

Mass flow rate of CH4 in air mixture = 700kg/h × (0.09CH4)/1 mix × (1/27.83kg/kmol) = 2.264 kmol/hr

Mass flow rate of air in mixture = 2.264kmol/h × 0.91kmol air/0.09kmolCH4 = 22.892 kmol/hr

We are told that the mixture is capable of being ignited if the mole percent of methane is between 5% and 15%.

Thus, for 5% of methane, the air required will be;

2.264kmol/h × 0.95kmol air/0.05kmol CH4 = 43.016 kmol/hr

Now, the dilution air needed will be =

43.016 - 22.892 = 20.124 kmol/hr

Total mass flow rate of mixture =

700kg/hr + (20.124kmol/hr × 29kg/mol) = 1283.596 kg/hr

We are told that air consist of 21 mole% Oxygen (O2).

Molar mass of oxygen = 32

Thus;

Mass fraction of oxygen in the product gas = 43.016kmol/h × (0.21molO2/1mol air) × (32kg oxygen/1kmol oxygen) × (1/1283.596kg/h) = 0.2252

Thus, written in percentage form, we have; 22.52%

So, percentage by mass of oxygen in the product gas = 22.52%

3 0

3 years ago

How much mercury and oxygen could be obtained from 21.7g of mercury (II) oxide

Musya8 [376]

20.06 g of Hg and 1.6 g of O₂

<u>Explanation:</u>

To Find:

Number of Mercury and oxygen that can be obtained from 21.7 g of HgO

First we have to write the balanced equation for the decomposition reaction of Mercury(II) oxide as,

2 HgO (s) → 2Hg(l) + O₂ (g)

21.7 g of HgO = $\frac{21.7 g}{216.59 g / mol}$

= 0.1 mol of HgO.

As per the above equation, we can find the mole ratio between HgO and Hg is 1: 1 and that of HgO and oxygen is 2:1 .

So amount of Hg produced = 0.1 mol × 200.59 g / mol ( molar mass of Hg)

= 20.06 g of Hg

Amount of oxygen produced = 0.05 mol × 32 g/ mol = 1.6 g of O₂

Thus it is clear that 20.06 g of Hg and 1.6 g of O₂ is obtained from 21.7 g of HgO

7 0

4 years ago

What factors do you think should be considered in selecting a metal to use to reclaim copper? Which factor do you think is the m

Lana71 [14]

Answer:

Aluminum

Explanation:

<h3> <em>Aluminum would be the best choice to reuse copper from the copper chloride solution because it In the aluminum–copper reaction, the chloride ions help remove the aluminum oxide coating so that a reaction can take place through the oxidation of aluminum and the reduction of copper ions.The aluminum–copper chloride reaction is a simplified version of a waste reclamation and reduction strategy for handling toxic-waste materials. </em></h3>

7 0

3 years ago

How many sodium atoms will create an ionic bond with sulfur?

Galina-37 [17]

Hi,

Two sodium atoms are needed to create an ionic bond with sulfur.

5 0

3 years ago