Answer:
electrolysis of brine
Explanation:
Rock salt deposits are usually mined; occasionally water is pumped down, and brine which contain 25 percent of sodium chloride is found
so d brine is electrolyzed to produce chlorine
The balanced equation for the reaction between KOH and HBr is as follows;
KOH + HBr --> KBr + H₂O
stoichiometry of KOH to HBr is 1:1
number of KOH moles reacted - 0.25 mol/L x 0.015 L = 0.00375 mol
according to molar ration
number of KOH moles reacted = number of HBr moles reacted
number of HBr moles reacted - 0.00375 mol
if 12 mL of HBr contains - 0.00375 mol
then 1000 mL of HBr contains - 0.00375 mol / 12 mL x 1000 mL = 0.313 mol
therefore molarity of HBr is 0.313 M
Number of moles of oxygen = x
number of moles of nitrogen = y
x = 2y
initial pressure, p1 = 0.8 atm
final pressure, p2 = 1.10 atm
At constant volume and temperature p1 / n1 = p2 / n2
=> p1 / p2 = n1 / n2
n1 = x + y = 2y + y = 3y
n2 = 0.2 + 3y
=> p1 / p2 = 3y / (0.2 + 3y)
=> 0.8 / 1.10 = 3y / (0.2 + 3y)
=> 0.8 (0.2 + 3y) = 1.10 (3y)
0.16 + 2.4y = 3.3y
=> 3.3y - 2.4y = 0.16
=> 0.9y = 0.16
=> y = 0.16 / 0.9
=. x = 2*0.16/0.9 = 0.356
Answer: 0.356 moles O2
52.8 grams of fat would be contained in 0.97 lbs of ground beef
Answer:
A = -213.09°C
B = 15014.85 °C
C = -268.37°C
Explanation:
Given data:
Initial volume of gas = 5.00 L
Initial temperature = 0°C (273 K)
Final volume = 1100 mL, 280 L, 87.5 mL
Final temperature = ?
Solution:
Formula:
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Conversion of mL into L.
Final volume = 1100 mL/1000 = 1.1 L
Final volume = 87.5 mL/1000 = 0.0875 L
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
T₂ = V₂T₁ / V₁
T₂ = 1.1 L × 273 K / 5.00 L
T₂ = 300.3 L.K / 5.00 K
T₂ = 60.06 K
60.06 K - 273 = -213.09°C
2)
V₁/T₁ = V₂/T₂
T₂ = V₂T₁ / V₁
T₂ = 280 L × 273 K / 5.00 L
T₂ = 76440 L.K / 5.00 K
T₂ = 15288 K
15288 K - 273 = 15014.85 °C
3)
V₁/T₁ = V₂/T₂
T₂ = V₂T₁ / V₁
T₂ = 0.0875 L × 273 K / 5.00 L
T₂ = 23.8875 L.K / 5.00 K
T₂ = 4.78 K
4.78 K - 273 = -268.37°C
