Answer:
Heat absorbed by water = 3985.26 j
Explanation:
Given data:
Mass of water = 75 g
Initial temperature = 20.0°C
Final temperature = 32.7°C
Specific heat of water = 4.184 j/g.°C
Heat absorbed by water = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 32.7°C - 20°C
ΔT = 12.7 °C
Q = 75 g ×4.184 j/g.°C ×12.7 °C
Q = 3985.26 j
Answer: ability to undergo a chemical reaction
Explanation:
Oxygen is used up and glucose is broken down.
Explanation:
Method of prepration of sodium thiosulphate - definition
In the laboratory, this salt can be prepared by heating an aqueous solution of sodium sulphite with sulphur or by boiling aqueous NaOH and sulfur according to this equation:

As mentioned above, phosphoric acid has 3 pKa values, and after 3 ionization it gives 3 types of ions at different pKa values:
H₃PO₄(aq)
+ H₂O(l) ⇌ H₃O⁺(aq) + H₂PO₄⁻ (aq) pKₐ₁
<span>
</span>H₂PO₄⁻(aq) + H₂O(l) ⇌ H₃O⁺(aq) + HPO₄²⁻ (aq) pKₐ₂
HPO₄²⁻(aq) + H₂O(l) ⇌ H₃O⁺(aq) + PO₄³⁻ (aq) pKₐ₃
At the highest pKa value (12.4) of phosphoric acid, the last OH group will lose its hydrogen. On the picture I attached, it is shown required protonated form of phosphoric acid before reaction whose pKa value is 12.4.