<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
The base gram of the compound is 16.042
To find the new amount of grams, you need to multiply the amount of moles by the the base grams
16.042*.25=4.01
So one fourth of the compound is 4.01 grams of Carbon tetrahydride.
Answer:Prophase is the first stage of mitosis. In prophase,
Explanation:
Answer:
K₃PO₄ + 3HCl ⇒ 3KCl + H₃PO₄
Explanation:
practice these, it gets easier. goal is to match the amount of reactants and products on both sides of the equation. cheers!
