<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
<h3><u>Answer;</u></h3>
1. In the light reactions, light energy is used to oxidize H2O to O2.
2. The electrons derived from this oxidation reaction in the light reactions are used to reduce NADP+ to NADPH.
3. The Calvin cycle oxidizes the light-reactions product NADPH to NADP+.
4. The electrons derived from this oxidation reaction in the Calvin cycle are used to reduce CO2 to G3P.
<h3><u>Explanation;</u></h3>
- <em><u>In the light reactions, light energy is used to remove electrons from (oxidize) water, producing O2 gas. These electrons are ultimately used to reduce NADP+ to NADPH.
</u></em>
- In the Calvin cycle, NADPH is oxidized back to NADP+ (which returns to the light reactions). The electrons released by the oxidation of NADPH are used to reduce three molecules of CO2 to sugar (G3P), which then exits the Calvin cycle.
- As ATP and NADPH are used in the Clavin cycle, they produce ADP and NADP+, respectively, which are returned to the light reactions so that more ATP and NADPH can be formed.
To solve this problem we will use the following equation:
w = (m of solute) / (m of solution)
w - percentage
It is necessary to mention here that mass of solution is a sum of the mass of solute and mass of water.
<span>w = mass CaCl2/(mass of water + mass of CaCl2)
</span>
mass of water = x
0.35 = 36 / (x + 36)
0.35 × (x + 36) = 36
0.35x + 12.6 = 36
0.35x = 23.4
x = 66.86 g of water is necessary
Answer:
see explanation below
Explanation:
To do this exercise, we need to use the following expression:
P = nRT/V
This is the equation for an ideal gas. so, we have the temperature of 22 °C, R is the gas constant which is 0.082 L atm / mol K, V is the volume in this case, 5 L, and n is the moles, which we do not have, but we can calculate it.
For the case of the oxygen (AW = 16 g/mol):
n = 30.6 / 32 = 0.956 moles
For the case of helium (AW = 4 g/mol)_
n = 15.2 / 4 = 3.8 moles
Now that we have the moles, let's calculate the pressures:
P1 = 0.956 * 0.082 * 295 / 5
P1 = 4.63 atm
P2 = 3.8 * 0.082 * 295 / 5
P2 = 18.38 atm
Finally the total pressure:
Pt = 4.63 + 18.38
Pt = 23.01 atm
When we have the balanced equation for this reaction:
AB3 ↔ A+3 + 3B-
So we can get Ksp:
when Ksp = [A+3][B-]^3
when [A+3] = 0.047 mol and from the balanced equation when
1 mol [A+3] → 3 mol [B-]
0.047 [A+3] → ??
[B-] = 3*0.047 = 0.141
so by substitution in Ksp formula:
∴Ksp = 0.047 * 0.141^3
= 1.32x10^-4
