1) Formulas:
a) mole fraction of component 1, X1
X1 = number of moles of compoent 1 / total number of moles
b) Molar mass = number grams / number of moles => number of moles = number of grams / molar mass
2) Application
Number of moles of CaI2 = 0.400
Molar mass of water = 18.0 g/mol
Number of moles of water: 850.0 g / 18.0 g/mol = 47.22 mol
Total number of moles = 0.400 + 47.22 =47.62
Molar fraction of CaI2 = 0.400 / 47.62 = 0.00840
Answer:
Element
Explanation:
Because elements can be made with only one atom the rest of the answers cant be
Answer:
1: [H+] = 0.01 M
2: [H+] = 0.0001 M
3: [H+] = 0.0001 M
Explanation:
Step 1: data given
pH = -log[H+]
pH = pOH = 14
Step 2:
1. A solution with pH = 2.0
pH = 2
-log[H+] = 2.0
[H+] = 10^-2
[H+] = 0.01 M
2. A solution with pH = 4.0
pH = 4
-log[H+] = 4.0
[H+] = 10^-4
[H+] = 0.0001 M
3. A solution with pOH = 10.0
pH = = 14 - 10 = 4
pH = 4
-log[H+] = 4.0
[H+] = 10^-4
[H+] = 0.0001 M
Answer:
C8H17N
Explanation:
Mass of the unknown compound = 5.024 mg
Mass of CO2 = 13.90 mg
Mass of H2O = 6.048 mg
Next, we shall determine the mass of carbon, hydrogen and nitrogen present in the compound. This is illustrated below:
For carbon, C:
Molar mass of CO2 = 12 + (2x16) = 44g/mol
Mass of C = 12/44 x 13.90 = 3.791 mg
For hydrogen, H:
Molar mass of H2O = (2x1) + 16 = 18g/mol
Mass of H = 2/18 x 6.048 = 0.672 mg
For nitrogen, N:
Mass N = mass of unknown – (mass of C + mass of H)
Mass of N = 5.024 – (3.791 + 0.672)
Mass of N = 0.561 mg
Now, we can obtain the empirical formula for the compound as follow:
C = 3.791 mg
H = 0.672 mg
N = 0.561 mg
Divide each by their molar mass
C = 3.791 / 12 = 0.316
H = 0.672 / 1 = 0.672
N = 0.561 / 14 = 0.040
Divide by the smallest
C = 0.316 / 0.04 = 8
H = 0.672 / 0.04 = 17
N = 0.040 / 0.04 = 1
Therefore, the empirical formula for the compound is C8H17N
