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Answer
% of oxygen(O) in glucose = 96/ 180 × 100 = 53.33
Explanation:
so the mass of oxygen in 1 mole of glucose = 6 × 16.00 g = 96.00 g. % oxygen = 96.00 × 100 % = 53.3 % by mass.
Potassium or any other metals.
Answer:
1008.0 kJ.
Explanation:
- Firstly, we need to calculate the no. of moles of 88.0 g of propane:
<em>n = mass/molar mass </em>= (88.0 g)(/(44.0 g/mol) = <em>2.0 mol.</em>
∴ The combustion of 2.0 moles of propane produces 4032.0 kJ.
<em><u>Using cross multiplication: </u></em>
The combustion of 2.0 moles of propane produces → 4032.0 kJ.
The combustion of 0.5 moles of propane produces → ??? kJ.
<em>∴ The amount of heat released for the combustion of 0.5 moles of propane </em>= (4032.0 kJ)(0.5 mol)/(2.0 mol) = <em>1008.0 kJ.</em>
Answer:
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