Answer:
25.7 kJ/mol
Explanation:
There are two heats involved.
heat of solution of NH₄NO₃ + heat from water = 0
q₁ + q₂ = 0
n = moles of NH₄NO₃ = 8.00 g NH₄NO₃ × 1 mol NH₄NO₃/80.0 g NH₄NO₃
∴ n = 0.100 mol NH₄NO₃
q₁ = n * ΔHsoln = 0.100 mol * ΔHsoln
m = mass of solution = 1000.0 g + 8.00 g = 1008.0 g
q₂ = mcΔT = 58.0 g × 4.184 J°C⁻¹ g⁻¹ × ((20.39-21)°C) = -2570.19 J
q₁ + q₂ = 0.100 mol ×ΔHsoln – 2570.19 J = 0
ΔHsoln = +2570.19 J /0.100 mol = +25702 J/mol = +25.7 kJ/mol
The compound's molecular formula is C2H6. This is obtained by:
mass moles divided by smallest moles
C 32g 32/12 = 2.67 1
H 8g 8/1.01 = 7.92 approx. 3
Next, divide both terms by the smallest number of moles, 2.67. This gives 1 and 3. So the empirical formula is CH3 which has a molar mass of 15g/mol. Given the molar mass of the molecular formula as 30g/mole, we can calculate the factor by which to multiply the subscripts of CH3.
X = molar mass of molecular formula / molar mass of empirical formula = 30/15
X=2
So (CH3)2 is C2H6.
Did you find the answer? Because I am on that question right now for my chem homework
Answer:easy
Explanation: the answer is '/.'/;==-.'/'.;.;pand'.[;';; because,/.;'[][]';;/..,./. hope it helps )
<h3>
Answer:</h3>
1.2 × 10^-11 M
<h3>
Explanation:</h3>
<u>We are given;</u>
Concentration of Hydrogen ions [H⁺] as 0.00083 M
<u>We are required to calculate the concentration of [OH⁻]</u>
We know that;
pH = - log [H⁺]
POH = -log[OH⁻]
Also, pH + pOH = 14
With the concentration of H⁺ we can calculate the pH
pH = -log 0.00083 M
= 3.08
But; pH + pOH = 14
Therefore; pOH = 14 - pH
= 14 - 3.08
= 10.92
But, pOH = -log[OH⁻]
Therefore; [OH] = -Antilog(pOH)
Hence; [OH⁻] = Antilog -10.92
= 1.2 × 10^-11 M
Therefore, [OH⁻] is 1.2 × 10^-11 M
