The heat used in phase changes is calculated by multiplying the mass of the substance by the energy of the phase change. In this case, for liquid to boil, we would find total heat by multiply the mass of liquid by the latent heat of vaporization (Hvap). If we are instead given the Hvap and the total heat of 1 kJ, we would divide 1 kJ by the Hvap (which is usually in kJ/kg) to get the mass of liquid boiled (in kg).
Answer:
D. Na+
Explanation:
Na+ is the only ion here that is not related to pH. Adding OH- increases pH, and adding H3O+ or H+ (the two are interchangeable) decreases pH.
Answer:
eight
Explanation:
Hence, from the above given configuration it is clear that the number of valence electrons in the Argon atom is eight.
Answer:
HNO₃
Explanation:
Data given
Nitrogen = 9.8 g
Hydrogen = 0.70 g
Oxygen = 33.6 g
Empirical formula = ?
Solution:
Convert the masses to moles
For Nitrogen
Molar mass of N = 14 g/mol
no. of mole = mass in g / molar mass
Put value in above formula
no. of mole = 9.8 g/ 14 g/mol
no. of mole = 0.7
mole of N = 0.7 mol
For Hydrogen
Molar mass of H = 1 g/mol
no. of mole = mass in g / molar mass
Put value in above formula
no. of mole = 0.70 g/ 1 g/mol
no. of mole = 0.7
mole of H = 0.7 mol
For Oxygen
Molar mass of O = 16 g/mol
no. of mole = mass in g / molar mass
Put value in above formula
no. of mole = 33.6 g / 16 g/mol
no. of mole = 2.1
mole of O = 2.1 mol
Now we have values in moles as below
N = 0.7
H = 0.7
O = 2.1
Divide the all values on the smallest values to get whole number ratio
N = 0.7 / 0.7 = 1
H = 0.7 / 0.7 = 1
O = 2.1 / 0.7 = 3
So all have following values
N = 1
H = 1
O = 3
So the empirical formula will be HNO₃ i.e. all three atoms in simplest small ratio.
Answer:
-2.80 × 10³ kJ/mol
Explanation:
According to the law of conservation of energy, the sum of the heat absorbed by the bomb calorimeter (Qcal) and the heat released by the combustion of the glucose (Qcomb) is zero.
Qcal + Qcomb = 0
Qcomb = - Qcal [1]
We can calculate the heat absorbed by the bomb calorimeter using the following expression.
Qcal = C × ΔT = 4.30 kJ/°C × (29.51°C - 22.71°C) = 29.2 kJ
where,
C: heat capacity of the calorimeter
ΔT: change in the temperature
From [1],
Qcomb = - Qcal = -29.2 kJ
The internal energy change (ΔU), for the combustion of 1.877 g of glucose (MW 180.16 g/mol) is:
ΔU = -29.2 kJ/1.877 g × 180.16 g/mol = -2.80 × 10³ kJ/mol
