<h3>
Answer:</h3>
28.96 kJ/°C
<h3>
Explanation:</h3>
We are given;
- Enthalpy change (ΔH) = −3226.7 kJ/mol
- The reaction is exothermic since the heat change is negative;
- Mass of benzoic acid = 3.1007 g
- Temperature change (21.84°C to 24.67°C) = 2.83°C
We are required to find the heat capacity of benzoic acid;
<h3>Step 1: Moles of benzoic acid </h3>
Moles = Mass ÷ molar mass
Molar mass of benzoic = 122.12 g/mol
Therefore;
Moles = 3.1007 g ÷ 122.12 g/mol
= 0.0254 moles
<h3>Step 2: Determine the specific heat capacity </h3>
Heat change for 1 mole = 3226.7 kJ
Moles of Benzoic acid = 0.0254 moles
But;
Specific heat capacity × ΔT = Moles × Heat change
cΔT = nΔH
Therefore;
Specific heat capacity,c = nΔH ÷ ΔT
= (3226.7 kJ × 0.0254 moles) ÷ 2.83°C
= 28.96 kJ/°C
Therefore, the specific heat capacity of benzoic acid is 28.96 kJ/°C
Answer : The temperature of the gas is, 245.9 K
Explanation :
To calculate the temperature of gas we are using ideal gas equation:

where,
P = pressure of gas = 2770.96 torr = 3.646 atm
Conversion used : (1 atm = 760 torr)
V = volume of gas = 88.84 L
T = temperature of gas = ?
R = gas constant = 0.0821 L.atm/mole.K
w = mass of gas = 609.64 g
M = molar mass of
gas = 38 g/mole
Now put all the given values in the ideal gas equation, we get:


Therefore, the temperature of the gas is, 245.9 K
Here, the three different notation of the p-orbital in different sub-level have to generate
The value of azimuthal quantum number (l) for -p orbital is 1. We know that the magnetic quantum number
depends upon the value of l, which are -l to +l.
Thus for p-orbital the possible magnetic quantum numbers are- -1, 0, +1. So there will be three orbitals for p orbitals, which are designated as
,
and
in space.
The three p-orbital can be distinguish by the quantum numbers as-
For 2p orbitals (principal quantum number is 2)
1) n = 2, l = 1, m = -1
2) n = 2, l = 1, m = 0
3) n = 2, l = 1, m = +1
Thus the notation of different p-orbitals in the sub level are determined.
Answer:
= 12 mL H202
Explanation:
Given that, the concentration of H2O2 is given antiseptic = 3.0 % v/v
It implies that, 3ml H2O2 is present in 100 ml of solution.
Therefore, to calculate the amount of H202 in 400.0 mL bottle of solution;
we have;
(3.0 mL/ 100 mL) × 400 mL
= 12 mL H202