The reaction between K₂SO₄(aq) and SrI₂(aq) produces KI(aq) and SrSO₄(s) as products.
The reaction is
K₂SO₄(aq) + SrI₂(aq) → KI(aq)+ SrSO₄(s)
To balance the equation both side of the reaction should have same number of atoms in each element.
Right hand side of the reaction has 1 K, 1 I, 1 Sr, 1 S and 4 O atoms while 2 K, 2 I, 1 Sr,1 S and 4 O present in left hand side of the reaction.
Hence, number of I atoms and number of K atoms are not balanced.
To balance the K atoms we should add 2 before KI. Then I atoms will be 2 at the right hand side.
Hence, the balanced reaction equation is
K₂SO₄(aq) + SrI₂(aq) → 2KI(aq)+ SrSO₄(s)
Answer:
The final balanced equation is :
Explanation:
Balancing in acidic medium:
First we will determine the oxidation and reduction reaction from the givne reaction :
Oxidation:
Balance the charge by adding 2 electrons on product side:
....[1]
Reduction :
Balance O by adding water on required side:
Now, balance H by adding on the required side:
At last balance the charge by adding electrons on the side where positive charge is more:
..[2]
Adding [1] and [2]:
The final balanced equation is :
The heat of the reaction, in kJ, when 4.18 g of the hydrocarbon are combusted 775.70 kJ.
The heat energy is given as :
q = m c ΔT + Ccal ΔT
q = ( 974 g× 4.184 ×6.9) + 624 ×6.9
q = 32424.59 J
moles of hydrocarbon = 0.0418 mol
heat of combustion = 32424.59 J / 0.0418 mol
= 775707.89 J
= 775.70 kJ
Thus, A 4.18 g sample of a hydrocarbon is combusted in a bomb calorimeter that contains 974 g of water. the temperature of the water increases by 6.9 °C when the hydrocarbon is combusted. the calorimeter constant for the calorimeter was determined to be 624 J/°C. what is the heat of the reaction is 775.70 kJ.
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CuO(aq) + H2(g) → H2O(l) + Cu(s)
CuO= 64 + 16= 80g
1 mole of CuO → 1 mole of Cu
80g of CuO → 1 mole of Cu
20.50g of CuO → y
y= 20.50/80= 0.26mole.