True I think I’m not for sure
Original molarity was 1.7 moles of NaCl
Final molarity was 0.36 moles of NaCl
Given Information:
Original (concentrated) solution: 25 g NaCl in a 250 mL solution, solve for molarity
Final (diluted) solution: More water is added to make the new total volume 1.2 liters, solve for the new molarity
1. Solve for the molarity of the original (concentrated) solution.
Molarity (M) = moles of solute (mol) / liters of solution (L)
Convert the given information to the appropriate units before plugging in and solving for molarity.
Molarity (M) = 0.43 mol NaCl solute / 0.250 L solution = 1.7 M NaCl (original solution)
2. Solve for the molarity of the final (diluted) solution.
Remember that the amount of solute remains constant in a dilution problem; it is just the total volume of the solution that changes due to the addition of solvent.
Molarity (M) = 0.43 mol NaCl solute / 1.2 L solution
Molarity (M) of the final solution = 0.36 M NaCl
I hope this helped:))
Answer:
Cu(NO3)2(aq)+Pb(s) ⇌ Pb(NO3)2(aq)+Cu(s)
Explanation:
If we look at the both reactions closely, we will quickly discover that the reaction CuSO4(aq)+Pb(s) ⇌ PbSO4(s)+Cu(s) involves PbSO4.
The compound PbSO4 is insoluble in water and sinks to the bottom of the reaction vessel. When this occurs, the concentration of Pb^2+ becomes low. This will bring about a low voltage in the cell.
On the other hand, Pb(NO3)2 is soluble in water hence the cell voltage in this case is higher than the former.
D. Drop in barometric pressure, warm ocean water, humid air. The low pressure brings in a cool air mass causing collision of two different masses.
My father rode out a typhoon near Okinawa WWII, onboard the battleship USS Missouri BB-63.
Violent pitching, alarms going off for approaching capsize pitch. The captain came on loudspeaker “ don’t worry men, land is near... about a mile straight down”.
Answer:
D) 5.15
Explanation:
Step 1: Write the equation for the dissociation of HCN
HCN(aq) ⇄ H⁺(aq) + CN⁻(aq)
Step 2: Calculate [H⁺] at equilibrium
The percent of ionization (α%) is equal to the concentration of one ion at the equilibrium divided by the initial concentration of the acid times 100%.
α% = [H⁺]eq / [HCN]₀ × 100%
[H⁺]eq = α%/100% × [HCN]₀
[H⁺]eq = 0.0070%/100% × 0.10 M
[H⁺]eq = 7.0 × 10⁻⁶ M
Step 3: Calculate the pH
pH = -log [H⁺] = -log 7.0 × 10⁻⁶ = 5.15
