2.4 mol NaSO2 × (6.02×10^23) f.units ÷ 1 mol
=1.44480000×10^25
Approximately 1.4×10^25, if you use sig.figs.
Answer:
368.92g
Explanation:
Firstly, let's balance the equation which is
2NO + O₂ ---> 2NO₂
Starting with 8.02 mol of NO let's calculate the moles of oxygen which is in a 2 : 1 molar ratio
2NO + O₂
2 : 1
8.02 mol : x mol
Moles of O₂ = 8.02 ÷ 2 = 4.01 mol
Doing the same thing for 18.75 mol of O₂ to calculate the number of moles of NO
2NO + O₂
2 : 1
x mol : 18.75 mol
Moles of NO = 18.75 × 2 = 37.5 however we are told we have 8.02 moles of NO, so we are unable to use 18.75 mol of O₂
Using 8.02 mol of NO to figure out the number of moles of NO₂ :
2NO : 2NO₂
They have the same molar ratio of 2 : 2, so the number of moles is 8.02
Using formula moles = mass / Molar mass
Rearranging to find mass = moles × molar mass
Molar mass of NO₂ = 14 + 16 + 16 = 46
Mass = 46 × 8.02 = 368.92g
Answer:
The correct answer is - 46.60 mL.
Explanation:
To find the volume of the gas at its new increased temperature we need to use Charl Law that shows the direct relationship between Volume and Temperature while Pressure remains constant.
V1 = 40 ml
T1 = 30 degree C + 273 = 303 K
V2 = ?
T2 = 80 degree C + 273 = 353 K
Charl Equation is:
V
1/T
1 = V
2/
T
2
(V1) * (T2)/ T1= V2
placing value:
40*353/303 = V2
= 14120/303
Vf = 46.60 mL
Answer:
6.25 g of Carbon is in Excess
Explanation:
The balance chemical equation is as follow:
C + O₂ → CO₂
Step 1: Calculate Moles of O₂ as:
Moles = Mass / M/Mass
Moles = 50 g / 32 g/mol
Moles = 1.562 mol of O₂
Step 2: Calculate Moles of C as;
Moles = 25 g / 12 g/mol
Moles = 2.08 mol
Step 3: Calculate the Limiting reagent:
According to equation,
1 moles of O₂ reacted with = 1 mole of C
So,
1.562 moles of O₂ will react with = X moles of C
Solving for X,
X = 1 mol × 1.562 mol / 1 mol
X = 1.562 mol of C
While we are provided with 2.08 mol of C. Means C is in excess.
Step 4: Calculate amount of Excess C as;
Excess Moles = Given Moles - Used Moles
Excess Moles = 2.08 mol - 1.562 mol
Excess Moles = 0.521 moles
Step 5: Converting Excess moles to Mass as:
Mass = Moles × M.Mass
Mass = 0.521 mol × 12 g/mol
Mass = 6.25 g