Answer:
D. an orbital notation of the atom
Explanation:
i took the test
Missing question: What is the vapor pressure of the solution at 25°<span>C?
n(NaCl) = 100 g </span>÷ 58,4 g/mol.
n(NaCl) = 1,71 mol.
NaCl → Na⁺ + Cl⁻, amount of ions are 2 · 1,71 mol = 3,42 mol.
n(CaCl₂) = 100 g ÷ 111 g/mol = 0,9 mol.
CaCl₂ → Ca²⁺ + 2Cl⁻, amount of ions 3 · 0,9 mol = 2,7 mol.
m(solution) = 1000 ml (1,00 L) · 1,15 g/ml = 1150 g.
m(H₂O) = 1150 g - 100 g - 100 g = 950 g.
n(H₂O) = 950 g ÷ 18 g/mol = 118,75 mol.
<span>water's mole fraction = 118,75 mol </span>÷ (118,75 mol + 2,7 mol + 3,42 mol).
water's mole fraction = 0,95.
p(solution) = 0,95 · 23 mmHg = 21,85 mmHg.
Answer:
The waitress thought the man meant H202 which is Hydrogen Peroxide. The man drank the Hydrogen Peroxide and then died.
You keep on saying join fast wdym?!
Answer:
Hydrogen: -141 kJ/g
Methane: -55kJ/g
The energy released per gram of hydrogen in its combustion is higher than the energy released per gram of methane in its combustion.
Explanation:
According to the law of conservation of the energy, the sum of the heat released by the combustion and the heat absorbed by the bomb calorimeter is zero.
Qc + Qb = 0
Qc = -Qb [1]
We can calculate the heat absorbed by the bomb calorimeter using the following expression.
Q = C . ΔT
where,
C is the heat capacity
ΔT is the change in the temperature
<h3>Hydrogen</h3>
Qc = -Qb = -C . ΔT = -(11.3 kJ/°C) . (14.3°C) = -162 kJ
The heat released per gram of hydrogen is:
![\frac{-162kJ}{1.15g} =-141 kJ/g](https://tex.z-dn.net/?f=%5Cfrac%7B-162kJ%7D%7B1.15g%7D%20%3D-141%20kJ%2Fg)
<h3>Methane</h3>
Qc = -Qb = -C . ΔT = -(11.3 kJ/°C) . (7.3°C) = -82 kJ
The heat released per gram of methane is:
![\frac{-82kJ}{1.50g} =-55kJ/g](https://tex.z-dn.net/?f=%5Cfrac%7B-82kJ%7D%7B1.50g%7D%20%3D-55kJ%2Fg)