Answer:
0.64 L
Explanation:
Step 1: Given data
Density of oil (ρ): 8.5 kg/LMass of oil (m): 5.4 kgVolume of oil (V): ?Step 2: Calculate the volume corresponding to 5.4 kg of oil
Density is an intrinsic property and it is equal to the quotient between the mass and the volume.
ρ = m/V
V = m/ρ
V = 5.4 kg/(8.5 kg/L) = 0.64 L
Which change represents an oxidation reaction?
Calculate the pH of each of the following aqueous solutions. (Enter your answers to two decimal places.) (a) 10.0 mL deionized water WebAssign will check your answer for the correct number of significant figures. 2.72 Incorrect: Your answer is incorrect. (b) 10.0 mL deionized water plus 5.0 mL of 0.10 M NaOH WebAssign will check your answer for the correct number of significant figures. (c) 10.0 mL deionized water plus 10.0 mL of 0.10 M NaOH WebAssign will check your answer for the correct number of significant figures. (d) 10.0 mL deionized water plus 15.0 mL of 0.10 M NaOH WebAssign will check your answer for the correct number of significant figures.
Answer:
a. pH = 7.0
b. pH = 12.52
c. pH = 12.70
d. pH = 12.78
Explanation:
a. Deionized water has the [H⁺] of pure water = 1x10⁻⁷ (Kw = 1x10⁻¹⁴ = [H⁺][OH⁻] - [H⁺] = [OH⁻ -)
pH = -log[H⁺] = 7
b. Moles NaOH = 5x10⁻³L * (0.10mol / L) = 5x10⁻⁴moles OH⁻ / 0.015L = 0.0333M = [OH⁻]
-Total volume = 10mL+5mL = 15mL = 0.015L
pOH = -log[OH⁻] = 1.48
pH = 14-pOH
pH = 12.52
c. Moles NaOH = 0.010L * (0.10mol / L) = 1x10⁻³moles OH⁻ / 0.020L = 0.0500M = [OH⁻]
-Total volume = 10mL+10mL = 20mL = 0.020L
pOH = -log[OH⁻] = 1.30
pH = 14-pOH
pH = 12.70
d. Moles NaOH = 0.015L * (0.10mol / L) = 1.5x10⁻³moles OH⁻ / 0.025L = 0.060M = [OH⁻]
-Total volume = 10mL+15mL = 25mL = 0.025L
pOH = -log[OH⁻] = 1.22
pH = 14-pOH
pH = 12.78