All categories

3 years ago

Using the equation PCl5(g)=PCl3(g)+Cl2(g), if PCl5 is removed, what will the equilibrium shift?

Chemistry

1 answer:

blondinia [14]3 years ago

5 0

The system will shift towards whats been removed, so equilibrium will shift towards reactants.

You might be interested in

What is the charge on an atom after it gains two electrons during the formation of a bond?

vekshin1

Answer:

two negative charges is the answer of your question

8 0

2 years ago

What is the freezing point (in degrees Celcius) of 4.14 kg of water if it contains 235.1 g of butanol, C 4 H 9 O H

Karolina [17]

Answer:

Explanation:

Molal freezing point depression constant of butanol Kf = 8.37⁰C /m

ΔTf = Kf x m , m is no of moles of solute per kg of solvent .

mol weight of butanol = 70 g

235.1 g of butanol = 235.1 / 70 = 3.3585 moles

3.3585 moles of butanol dissolved in 4.14 kg of water .

ΔTf = 8.37 x 3.3585 / 4.14

= 6.79⁰C

Depression in freezing point = 6.79

freezing point of solution = - 6.79⁰C .

5 0

2 years ago

Which of the drawings contain a mixture?

Assoli18 [71]

the answer is c

Explanation:

because c has all elements mixed together

4 0

3 years ago

How many moles of k3po4 can be formed when 4.4 moles of h3po4 react with 3.8 moles of koh? h3po4 + koh yields h2o + k3po4 be sur

schepotkina [342]

the balanced chemical equation for the reaction is as follows

H₃PO₄ + 3KOH ---> K₃PO₄ + 3H₂O

stoichiometry of H₃PO₄ to KOH is 1:3

first we have to find which the limiiting reactant is

as the amount of product formed depends on the amount of limiting reactant present

number of H₃PO₄ moles reacted - 4.4 mol

if H₃PO₄ is the limiting reactant

1 mol of H₃PO₄ reacts with 3 mol of KOH

then 4.4 mol of H₃PO₄ reacts with - 3 x 4.4 mol = 13.2 mol of KOH

but only 3.8 mol of KOH is present

therefore KOH is the limiting reactant

stoichiometry of KOH to K₃PO₄ is 3:1

number of KOH moles reacted - 3.8 mol

therefore number of K₃PO₄ formed = number of KOH moles reacted / 3

= 3.8 mol / 3 = 1.3 mol

answer is 1.3 mol of K₃PO₄

3 0

3 years ago

A gas has a pressure of 810 kPa at 227°C. What will its pressure be at 27°C?

NeX [460]

486 kPa will be the pressure of gas at 27 degrees.

Explanation:

Data given :

Initial pressure of the gas P1 = 810 kPa

initial temperature of the gas T1 = 227 degrees OR 500.15 K

final pressure of the gas =?

final temperature of the gas = 27 degrees OR 300.15 K

Gay Lussac's law is used to calculate the pressure of the gas at 27 degrees or 300.15 K

$\frac{P1}{T1}$ = $\frac{P2}{T2}$

P2 = $\frac{P1T2}{T1}$

Putting the values in the equation:

P2 = $\frac{810 X 300.15}{500.15}$

P2 = 486 KPa

486 Kpa is the pressure of the gas when temperature was reduced to 27 degrees.

5 0

3 years ago