Answer:
React it hydrogen bromide
Step-by-step explanation:
Alkenes are <em>hydrohalogenated</em> by addition of HCl or HBr.
The equation for the reaction is
CH₂=CH₂ + HBr ⟶ CH₃CH₂Br
Here is the correct answer of the given question above about the crucible. The crucible and its contents gain mass when heated because <span>it contains a compound that forms a stable and solid oxide. Hope this is the answer that you are looking for. Have a nice day!</span>
Answer:
Volume is 1.065L
Explanation:
Hello,
We can easily solve this problem by using general gas equation.
PV / T = K
P1V1/T1 = P2V2/T2
Data;
P1 = 3.0atm
P2 = 1.0atm
T1 = 20°C = (20 + 273.15)K = 293.15K
T2 = 20°C = (20 + 273.15)K = 293.15K
V1 = 355mL = 0.355L
V2 = ?
From the data given, we can substitute it into the equation,
(P1 × V1) / T1 = (P2 × V2) / T2
(3.0 × 0.355) / 293.15 = (1.0 × V2) / 293.15
1.065 = 1.0V2
Divide both sides by 1.0
V2 = 1.065L
The volume of CO₂ released is 1.065L
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⁻]
<em>-Total volume = 10mL+5mL = 15mL = 0.015L</em>
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⁻]
<em>-Total volume = 10mL+10mL = 20mL = 0.020L</em>
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⁻]
<em>-Total volume = 10mL+15mL = 25mL = 0.025L</em>
pOH = -log[OH⁻] = 1.22
pH = 14-pOH
pH = 12.78
A is the correct answer i really hope this helps!