Volume fraction = volume of the element / volume of the alloy
Volume = density * mass
Base: 100 grams of alloy
mass of tin = 15 grams
mass of lead = 85 grams
volume = mass / density
Volume of tin = 15g / 7.29 g/cm^3 = 2.06 cm^3
Volume of lead = 85 g / 11.27 g/cm^3 = 7.54 cm^3
Volume fraction of tin = 2.06 cm^3 / (2.06 cm^3 + 7.54 cm^3) = 0.215
Volume fraction of lead = 7.54 cm^3 / (2.06 cm^3 + 7.54 cm^3) = 0.785
As you can verify the sum of the two volume fractions equals 1: 0.215 + 0.785 = 1.000
Answer:
number of moles of NaCl produce = 12 mol
Explanation:
Firstly, we need to write the chemical equation of the reaction and balance it .
Na(s) + Cl2(g) → NaCl(s)
The balanced equation is as follows:
2Na(s) + Cl2(g) → 2NaCl(s)
1 mole(71 g) of chlorine produces 2 moles(117 g) of sodium chloride
6 mole of chlorine gas will produce ? mole of sodium chloride
cross multiply
number of moles of NaCl produce = 6 × 2
number of moles of NaCl produce = 12 moles
number of moles of NaCl produce = 12 mol
Answer:
= 9.28 g CO₂
Explanation:
First write a balanced equation:
CH₄ + 2O₂ -> 2H₂O + CO₂
Convert the information to moles
7.50g CH₄ = 0.46875 mol CH₄
13.5g O₂ = 0.421875 mol O₂
Theoretical molar ratio CH₄:O₂ -> 1:2
Actual ratio is 0.46875 : 0.421875 ≈ 1:1
If all CH₄ is used up, there would need to be more O₂
So O₂ is the limiting reactant and we use this in our equation
Use molar ratio to find moles of CO₂
0.421875 mol O₂ * 1 mol CO₂/2 mol O₂=0.2109375 mol CO₂
Then convert to grams
0.2109375 mol CO₂ = 9.28114 g CO₂
round to 3 sig figs
= 9.28 g CO₂
Answer:
1. 43.44g of HCl
2. 26.67 L of HCl
Explanation:
1) Molarity of a solution = number of moles (n) ÷ Volume (V)
According to the provided information in this question,
V = 350 mL = 350/1000 = 0.350L
Molarity = 3.4 M
Using Molarity = n/V
3.4 = n/0.350
n = 3.4 × 0.350
n = 1.19mol
Using the formula below to calculate the mass of HCl;
mole = mass/molar mass
Molar mass of HCl = 1 + 35.5 = 36.5g/mol
mole = mass/MM
mass = 1.19 mol × 36.5g/mol
mass = 43.44g of HCl
2) At STP, HCl has a pressure of 1atm, a temperature of 273K
V = ?
n = 1.19 mol
R = 0.0821 Latm/molK
Using PV = nRT
V = nRT/P
V = 1.19 × 0.0821 × 273/1
Volume = 26.67L
