I am pretty sure the answer is B but correct me if I'm wrong hope this helps.
Answer is: adding NaCl will lower the freezing point of a solution.
A solution (in this example solution of sodium chloride) freezes at a lower temperature than does the pure solvent (deionized water).
The higher the solute concentration (sodium chloride), freezing point depression of the solution will be greater.
Equation describing the change in freezing point:
ΔT = Kf · b · i.
ΔT - temperature change from pure solvent to solution.
Kf - the molal freezing point depression constant.
b - molality (moles of solute per kilogram of solvent).
i - Van’t Hoff Factor.
Dissociation of sodium chloride in water: NaCl(aq) → Na⁺(aq) + Cl⁻(aq).
Answer:
posotive
Explanation:
electrons give off negative pulse so take one away
Answer: 12.78ml
Explanation:
Given that:
Volume of KOH Vb = ?
Concentration of KOH Cb = 0.149 m
Volume of HBr Va = 17.0 ml
Concentration of HBr Ca = 0.112 m
The equation is as follows
HBr(aq) + KOH(aq) --> KBr(aq) + H2O(l)
and the mole ratio of HBr to KOH is 1:1 (Na, Number of moles of HBr is 1; while Nb, number of moles of KOH is 1)
Then, to get the volume of a 0.149 m potassium hydroxide solution Vb, apply the formula (Ca x Va)/(Cb x Vb) = Na/Nb
(0.112 x 17.0)/(0.149 x Vb) = 1/1
(1.904)/(0.149Vb) = 1/1
cross multiply
1.904 x 1 = 0.149Vb x 1
1.904 = 0.149Vb
divide both sides by 0.149
1.904/0.149 = 0.149Vb/0.149
12.78ml = Vb
Thus, 12.78 ml of potassium hydroxide solution is required.
Answer:
5.37 L
Explanation:
To solve this problem we need to use the PV=nRT equation.
First we <u>calculate the amount of CO₂</u>, using the initial given conditions for P, V and T:
- P = 785 mmHg ⇒ 785/760 = 1.03 atm
- T = 18 °C ⇒ 18 + 273.16 = 291.16 K
1.03 atm * 4.80 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 291.16 K
We <u>solve for n</u>:
Then we use that value of n for another PV=nRT equation, where T=37 °C (310.16K) and P = 745 mmHg (0.98 atm).
- 0.98 atm * V = 0.207 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 310.16 K
And we <u>solve for V</u>:
