Answer:
Explanation:Are You From Milo?
Answer : The molar mass of the solute will be
87.90 g/mol.Explanation : We know the formula for elevation in boiling point, which is
Δt = i

m
given that, Δt = 0.357,

= 5.02 and mass of

= 40,
on substituting the value we get,
0.357 = (1) X (5.02) X (x/ 0.044), on solving we get x = 2.844 X

.
Now, 0.250/ 2.844 X

=
87.90 g/mol. which is the weight of unknown component.
The equilibrium reaction, causes the water dissociation constant, Kw, is 1.01 × 10-14<span> at 25 °C. That is because every H</span>+<span> (H</span>3O+) ion these forms accompanied by the formation of an OH-<span> ion, are the concentrations of these ions and in pure water the same thing can be calculated from </span>Kw<span>.
HOPED THIS HELP OUT ;)
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Answer:
2022 L
Explanation:
Ideal gas laws will work for gas in the balloon
The general gas law is for a gas at two arbitrary states 1 and 2 is given by
(P₁ V₁)/T₁ = (P₂ V₂)/T₂
P₁ = 1.17 atm
V₁ = 200.0 L
T₁ = 20°C = 293.15 K
P₂ = 63 mmHg = 0.0829 atm
V₂ = ?
T₂ = 210 K
(1.17 × 200)/293.15 = (0.0829 × V₂)/210
V₂ = (210 × 1.17 × 200)/(293.15 × 0.0829)
V₂ = 2022 L
Answer : The volume of 4.9 M
stock solution used to prepare the solution is, 12.24 ml
Solution : Given,
Molarity of aqueous
solution = 1.20 M = 1.20 mole/L
Volume of aqueous
solution = 50.0 ml = 0.05 L
(1 L = 1000 ml)
Molarity of
stock solution = 4.9 M = 4.9 mole/L
Formula used :

where,
= Molarity of aqueous
solution
= Molarity of
stock solution
= Volume of aqueous
solution
= Volume of
stock solution
Now put all the given values in this formula, we get the volume of
stock solution.

By rearranging the term, we get

Therefore, the volume of 4.9 M
stock solution used to prepare the solution is, 12.24 ml