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-Dominant- [34]

2 years ago

12

2. Suggest four ways in which the concentration of PH3 could be increased in an equilibrium described by the following equation:

P4(g)+6H2(g) ⇌ 4PH3(g) ΔH=110.5kJ

1 answer:

Nesterboy [21]2 years ago

4 0

Answer

increase in temperature

decrease in pressure

continuous removal of PH3
adding more of P into the system

Explanation:

In the reaction P4(g)+6H2(g) ⇌ 4PH3(g);

The effect of temperature on equilibrium has to do with the heat of reaction. Recall that for an endothermic reaction, heat is absorbed in the reaction, and the value of ΔH is positive. Thus, for an endothermic reaction, we can picture heat as being a reactant:

heat+A⇌BΔH=+

Since the reaction is endothermic reaction, heat is a absorbed. Decreasing the temperature will shift the equilibrium to the left, while increasing the temperature will shift the equilibrium to the right forming more of PH3.

According to Le Chatelier’s principle, adding additional reactant to a system will shift the equilibrium to the right, towards the side of the products. In the same Way, reducing the concentration of the product will also shift equilibrium to the right continually forming PH3 as it is removed.

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The volume of a gas is decreased from 100 liters at 173.0°C to 50 liters at a constant pressure. After the decrease in volume, w

Minchanka [31]

Answer:

223.08 K

Explanation:

First we <u>convert 173.0 °C to K</u>:

173.0 °C + 273.16 = 446.16 K

With the absolute temperature we can use <em>Charles' law</em> to solve this problem:

T₁V₂=T₂V₁

Where in this case:

T₁ = 446.16 K

V₂ = 50 L

T₂ = ?

V₁ = 100 L

We <u>input the data</u>:

446.16 K * 50 L = T₂ * 100 L

And <u>solve for T₂</u>:

T₂ = 223.08 K

5 0

1 year ago

Menthol is a flavoring agent extracted from peppermint oil. It contains C, H, and O. In one combustion analysis, 10.00 mg of the

RideAnS [48]

Answer:

C₁₀H₂₀O

Explanation:

The molecular formula must be $C_{x}H_yOz$ . The combustion reaction will occur between the fuel and oxygen gas:

$C_{x}H_yOz$ + O₂ → CO₂ + H₂O

For Lavoisier law, the mass of the reactants must be equal to the mass of the products (mass conservation):

10.0 + mO₂ = 28.16 + 11.53

mO₂ = 29.69 mg

Supposing that all the oxygen and the menthol were consumed, let's calculate the number of moles of the compounds, knowing, for the Periodic Table, that:

MC = 12 g/mol, MO = 16 g/mol, MH = 1 g/mol

MCO₂ = 12 + 2x16 = 44 g/mol

MH₂O = 2x1 + 16 = 18 g/mol

MO₂ = 2x16 = 32 g/mol

n = mass (g)/molar mass

nCO₂ = 0.02816/44 = 6.4x10⁻⁴ mol

nH₂O = 0.01153/18 = 6.4x10⁻⁴ mol

nO₂ = 0.02969/32 = 9.3x10⁻⁴ mol

The molar number is proportional in the molecule, so, in CO₂, the number of C is 6.4x10⁻⁴ mol, and of O is 1.28x10⁻³. All the carbon of the methol will be in CO₂, and all the H will be in H₂O. The number of moles of O will be the difference of moles in H₂O and the O₂, then:

nC = 6.4x10⁻⁴ mol

nH = 2x6.4x10⁻⁴ = 1.28x10⁻³ mol

nO = (2x6.4x10⁻⁴ + 6.4x10⁻⁴) - (2x9.3x10⁻⁴) = 6.0x10⁻⁵

The empirical formula is the molecule formula with the small subscripts numbers, which represent the number of moles of the atoms in the molecule. So, let's divide all the number of moles for the small on 6.0x10⁻⁵.

nC = (6.4x10⁻⁴)/(6.0x10⁻⁵) = 10

nH = (1.28x10⁻³)/(6.0x10⁻⁵) = 20

nO = (6.0x10⁻⁵)/(6.0x10⁻⁵) = 1

So, the empirical formula of methol is C₁₀H₂₀O.

3 0

2 years ago

A compound is 2.00% H by mass, 32.7% S by mass, and 65.3% O by mass. What is its empirical formula? The second step is to calcul

Free_Kalibri [48]

Lets take 100 g of this compound,
so it is going to be 2.00 g H, 32.7 g S and 65.3 g O.

2.00 g H *1 mol H/1.01 g H ≈ 1.98 mol H
32.7 g S *1 mol S/ 32.1 g S ≈ 1.02 mol S
65.3 g O * 1 mol O/16.0 g O ≈ 4.08 mol O

1.98 mol H : 1.02 mol S : 4.08 mol O = 2 mol H : 1 mol S : 4 mol O

Empirical formula
H2SO4

8 0

2 years ago

Read 2 more answers

Which of the following types of molecules always has a dipole moment? Linear molecules with two identical bonds. Trigonal pyrami

natali 33 [55]

Answer:

Trigonal pyramid molecules (three identical bonds)

Explanation:

In trigonal pyramidal molecule like molecule of ammonia , the vector some of intra- molecular dipole moment is not zero because the bonds are not symmetrically oriented . In other molecules , bonds are symmetrically oriented in space so the vector sum of all the internal dipole moment vectors cancel each other to make total dipole moment zero.

8 0

2 years ago

A quantity of 85.0 mL of 0.900 M HCl is mixed with 85.0 mL of 0.900 M KOH in a constantpressure calorimeter that has a heat capa

bogdanovich [222]

Explanation:

The given data is as follows.

$V_{1}$ = 85.0 ml, $M_{1}$ = 0.9 M

$V_{2}$ = 85.0 ml, $M_{1}$ = 0.9 M

Hence, number of moles of HCl and KOH will be the same because both the solutions have same volume and molarity.

So, No. of moles = Molarity × Volume

= $0.9 M \times 0.085 L$ (as 1 L = 1000 ml so, 85 ml = 0.085 L)

= 0.076 mol

As 1 mole gives 56.2 kJ/mol of heat of neutralization. Hence, calculate the heat of neutralization given by 0.076 moles as follows.

$56.2 kJ/mol \times 0.076 mol$

= 4.271 kJ

or, = 4271 J (as 1 kJ = 1000 J)

Therefore, heat released = - heat of gained by calorimeter

Since, it is given that density of the solution is similar to the density of water which is 1 g/ml.

Hence, mass of HCl = density × Volume of HCl

= 1.00 g/ml × 85.0 ml

= 85 g

Similarly, mass of KOH = = density × Volume of HCl

= 1.00 g/ml × 85.0 ml

= 85 g

Hence, total mass of the solution = 85 g + 85 g

= 170 g

Also, q = $mC \Delta T$

4271 J = $170 g \times 325 J/^{o}C \times (T_{f} - 18.24)^{o}C$

0.0773 = $T_{f} - 18.24$

$T_{f}$ = $18.317^{o}C$

Thus, we can conclude that final temperature of the mixed solution is $18.317^{o}C$ .

6 0

2 years ago