Answer:
M(Fe₂O₃) = 159.70 g/mol
M(CO) = 28.01 g/mol
M(Fe) = 55.85 g/mol
M(CO₂) = 44.01 g/mol
Explanation:
We can calculate the molar mass of a compound by summing the molar masses of the elements that form it.
Fe₂O₃
M(Fe₂O₃) = 2 × M(Fe) + 3 × M(O) = 2 × 55.85 g/mol + 3 × 16.00 g/mol = 159.70 g/mol
CO
M(CO) = 1 × M(C) + 1 × M(O) = 1 × 12.01 g/mol + 1 × 16.00 g/mol = 28.01 g/mol
Fe
M(Fe) = 1 × M(Fe) = 1 × 55.85 g/mol = 55.85 g/mol
CO₂
M(CO₂) = 1 × M(C) + 2 × M(O) = 1 × 12.01 g/mol + 2 × 16.00 g/mol = 44.01 g/mol
To solve this, we simply equate the change in enthalpy for
the two substances since heat gained by water is equal to heat lost of aluminum.
We know that the heat capacity of aluminum is 0.089 J/g°C and that of water is
4.184 J/g°C. Therefore:
450.2 (95.2 - T) (0.089) = 60 (T – 10) (4.184)
3,814.45456 – 40.0678 T = 251.04 T – 2,510.4
291.1078 T = 6,324.85456
<span>T = 21.7°C</span>
Answer:
The Ka is 9.11 *10^-8
Explanation:
<u>Step 1: </u>Data given
Moles of HX = 0.365
Volume of the solution = 835.0 mL = 0.835 L
pH of the solution = 3.70
<u>Step 2:</u> Calculate molarity of HX
Molarity HX = moles HX / volume solution
Molarity HX = 0.365 mol / 0.835 L
Molarity HX = 0.437 M
<u />
<u>Step 3:</u> ICE-chart
[H+] = [H3O+] = 10^-3.70 = 1.995 *10^-4
Initial concentration of HX = 0.437 M
Initial concentration of X- and H3O+ = 0M
Since the mole ratio is 1:1; there will react x M
The concentration at the equilibrium is:
[HX] = (0.437 - x)M
[X-] = x M
[H3O+] = 1.995*10^-4 M
Since 0+x = 1.995*10^-4 ⇒ x=1.995*10^-4
[HX] = 0.437 - 1.995*10^-4 ≈ 0.437 M
[X-] = x = 1.995*10^-4 M
<u>Step 4: </u>Calculate Ka
Ka = [X-]*[H3O+] / [HX]
Ka = ((1.995*10^-4)²)/ 0.437
Ka = 9.11 *10^-8
The Ka is 9.11 *10^-8
Answer:
Water has a lot of physical properties.
Explanation:
One of them is heat of vaporization.
Bam
