When we convert the given mass in grams and volume in liters to m/v percent, we recall that m/v percent is expressed as grams/100 milliliters. In this case the expression becomes (50 grams/ 2500 L)*(0.1L/100ml), that is equal to 0.002 grams/ 100 mL. Hence the the concentration is equal to 0.2 m/v percent.
Heat produced = -13588.956 kJ
<h3>Further explanation</h3>
Given
The reaction of combustion of Methane
CH4(g)+2O2(g)→CO2(g)+2H2O(g) ΔH∘rxn=−802.3kJ
271 g of CH4
Required
Heat produced
Solution
mol of 271 g CH₄ (MW=16 g/mol0
mol = mass : MW
mol = 271 : 16
mol = 16.9375
So Heat produced :
= mol x ΔH°rxn
= 16.9375 mol x −802.3kJ/mol = -13588.956 kJ
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Answer:
2PbSO4 → 2PbSO3 + O2
Explanation:
in original equation we notice that we have one extra oxygen, which we cannot form a O2 with, so by multiplying everything else by 2, we get 2 extra oxygen
The formula for K+ and S-2 is K2S because you need 2 potassium ions to balance out the sulfide ion
