Answer: 770 g water are needed to dissolve 27.8 g of ammonium nitrate
in order to prepare a 0.452 m solution
Explanation:
Molality : It is defined as the number of moles of solute present per kg of solvent
Formula used :

where,
n= moles of solute
Moles of
= weight of the solvent in g = ?


Thus 770 g water are needed to dissolve 27.8 g of ammonium nitrate
in order to prepare a 0.452 m solution
The correct Lewis structure for Fluorine is A.
Answer:The formulas of ionic compounds are:
a)
b)
c)
d)
Explanation:
Formulas for the an ionic compounds is determine by:
Criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.
(a) Copper bromide :Given that it contains
ion.

(b) Manganese oxide : Given that it contains
ion.

(c)Mercury iodide :Given that it contains 

(d) Magnesium phosphate :Given that it contains 

Answer: The workdone W = 505J
Explanation:
Applying the pressure-volume relationship
W= - PΔV
Where negative sign indicates the power is being delivered to the surrounding
W = - 1.0atm * ( 5.88 - 0.9)L
= - 1.0atm * (4.98)
W = -4.98 atmL
Converting to Joules
1atmL = 101.325J
-4.98atmL = x joules.
Work done in J = -4.98 * 101.325
W= -505J
Therefore the workdone is -505J
Complete question:
Consider the reaction.
At equilibrium at 600 K, the concentrations are as follows.
2HF -----> H₂ + F₂
[HF] = 5.82 x 10-2 M
[H2] = 8.4 x 10-3 M
[F2] = 8.4 x 10-3 M
What is the value of Keq for the reaction expressed in scientific notation?
2.1 x 10-2
2.1 x 102
1.2 x 103
1.2 x 10-3
Answer:
2.1 × 10^-2
Explanation:
Kequilibrum(Keq) = product/reactant
Equation for the reaction :
2HF -----> H₂ + F₂
Therefore,
Keq = [H2][F2] / [HF]^2
Keq = [8.4 x 10-3][8.4 x 10-3] / [5.82 x 10-2]^2
Keq = [70.56 × 10^(-3 + - 3)]/[33.8724 × 10^(-2×2)]
Keq = [70.56 × 10^-6] / [33.8724 × 10^-4]
Keq = 2.0665 × 10^(-6 - (-4))
Keq = 2.0665 × 10^(-6 + 4)
Keq = 2.1 × 10^-2