Answer:
1.17 grams of HCl can neutralize 2.7 grams sodium bicarbonate
Explanation:
Step 1: Data given
Mass of sodium bicarbonate = 2.7 grams
Step 2: The balanced equation
HCl + NaHCO3 ⇔ NaCl + H2O + CO2
Step 3: Calculate moles NaHCO3
moles NaHCO3 =2.7 g / 84 g/mol= 0.032 moles
Step 4: Calculate moles HCl
For 1 mol NaHCO3 we need 1 mol HCl
For 0.032 moles NaHCO3 = 0.032 moles HCl
Step 5: Calculate mass HCl
Mass HCl = moles HCl * molar mass HCl
mass HCl = 0.032 * 36.46 g/mol= 1.17 grams
1.17 grams of HCl can neutralize 2.7 grams sodium bicarbonate
Answer:
V₂ = 3227.46 L
Explanation:
Given data:
Initial volume of gas = 1000 L
Initial temperature = 50°C (50 +273 = 323 K)
Initial pressure = 101.3 KPa
Final pressure = 27.5 KPa
Final temperature = 10°C (10 +273 = 283 K)
Final volume = ?
Solution:
According to general gas equation:
P₁V₁/T₁ = P₂V₂/T₂
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₂ = P₁V₁ T₂/ T₁ P₂
V₂ = 101.3 KPa × 1000 L × 283 K / 323 K × 27.5 KPa
V₂ = 28667900 KPa .L. K /
8882.5 K.KPa
V₂ = 3227.46 L
Makes zero sense. What’s the question?
The answer would be 1,3,1,3
Answer:
A) 0.95 mol
Explanation:
We will assume the gas given off in the fermentation is an ideal gas because that allows us to use the ideal gas equation.
PV = nRT
First let's convert all measurements to units that we can use
P = 702 mmHg * 1 atm/760 mmHg = 0.92368 atm
V = 25.0 L
R = 0.08206 L-atm/mol-K
T = 22.5 °C +273.15 = 295.65 K
PV = nRT
0.92368 atm * 25.0 L = n * 0.08206 L-atm/mol-K * 295.65 K
n = 0.9518 mol
