# Balance the following: ___ AlBr3+ ___ K---> ___KBr+ ___ Al

Answer: ___AlBr3 + ___K -> ___KBr + ___ Al

1 AlBr3 + 3K -> 3KBr + 1 Al

hope this helps............

## Related Questions

Which statement must be true about a chemical system at equilibrium? (1) The forward and reverse reactions stop.
(2) The concentration of reactants and products are equal.
(3) The rate of the forward reaction is equal to the rate of the reverse reaction.
(4) The number of moles of reactants is equal to the number of moles of product.

(1) is wrong because at equilibrium both the reactant and the product are still being created they are just being created at the same rate meaning that

(3) is the answer. As the concentrations do not change, but the solution/system is still in dynamic equilibrium.
I agree with the answer below (partial credit would go to her/him I guess)
My answer I guess is more visual to explain
1 says --> |   which is true because that's the def of Equilibrium

In a mixture of He, Ne, and Xe gases with a total pressure of 925 atm, if there is 10.5 g of each gas in the mixture, what is the partial pressure of Xe?

Answer is: partial pressure of Xe is 22,95 atm.
m(He) = 10,5 g.
n(He) = m(He) ÷ M(He).
n(He) = 10,5 g ÷ 4 g/mol
n(He) = 2,625 mol.
m(Ne) = 10,5 g.
n(Ne) = m(Ne) ÷ M(Ne).
n(Ne) = 10,5 g ÷ 20,18 g/mol.
n(Ne) = 0,52 mol.
m(Xe) = 10,5 g.
n(Xe) = 10,5 g ÷ 131,3 g/mol.
n(Xe) = 0,08 mol.
Using Dalton's law:
p(Xe) = (0,08 mol / 0,08 mol + 0,52 mol + 2,625 mol) · 925 atm.
p(Xe) =  22,95 atm.

Partial pressure of Xe = 22.9 atm

Explanation:

Given:

Total pressure of gas mixture = 925 atm

Mass of He = Ne = Xe = 10.5 g

To determine:

Partial pressure of Xe

Explanation:

As per Dalton's Law in a mixture of gases the total pressure is the equal to the sum of partial pressures

In this case:

-------------(1)

Partial pressure of each gas can be expressed as:

-----(2)

where n(Xe) = moles of Xe

n(Total) = total moles

Therefore,

n(Total) = 0.080+2.625+0.525 = 3.23

Substituting for nXe, n(Total) and P(total) in equation (2)

Why is vigorous trituration required to prepare emulsions?