<em>I</em><em> </em><em>do</em><em> </em><em>not</em><em> </em><em>understand</em><em> </em><em>science</em><em> </em><em>but</em><em> </em><em>if</em><em> </em><em>u</em><em> </em><em>ask</em><em> </em><em>me</em><em> </em><em>I</em><em> </em><em>would</em><em> </em><em>have</em><em> </em><em>no</em><em> </em><em>clue</em><em> </em><em>do</em><em> </em><em>u</em><em> </em><em>get</em><em> </em><em>what</em><em> </em><em>I</em><em> </em><em>mean</em>
Answer:
30%
Explanation:
<em>This is the chemical formula for zinc bromate: Zn(BrO₃)₂. Calculate the mass percent of oxygen in zinc bromate. Round your answer to the nearest percentage.</em>
Step 1: Determine the mass of 1 mole of Zn(BrO₃)₂
M(Zn(BrO₃)₂) = 1 × M(Zn) + 2 × M(Br) + 6 × M(O)
M(Zn(BrO₃)₂) = 1 × 65.38 g/mol + 2 × 79.90 g/mol + 6 × 16.00 g/mol
M(Zn(BrO₃)₂) = 321.18 g/mol
Step 2: Determine the mass of oxygen in 1 mole of Zn(BrO₃)₂
There are 6 moles of atoms of oxygen in 1 mole of Zn(BrO₃)₂.
6 × m(O) = 6 × 16.00 g = 96.00 g
Step 3: Calculate the mass percent of oxygen in Zn(BrO₃)₂
%O = mO/mZn(BrO₃)₂ × 100%
%O = 96.00 g/321.18 g × 100% ≈ 30%
Answer:
yes it should react yoyoyoyoyo
The mass of CO₂ gas produced during the combustion of one gallon of octane is 8.21 kg.
The given parameters:
- <em>Density of the octane, ρ = 0.703 g/ml</em>
- <em>Volume of octane, v = 3.79 liters</em>
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The mass of the octane burnt is calculated as follows;
The combustion reaction of octane is given as;
From the reaction above:
228.46 g of octane -------------------> 704 g of CO₂ gas
2,664.37 of octane --------------------> ? of CO₂ gas
Thus, the mass of CO₂ gas produced during the combustion of one gallon of octane is 8.21 kg.
Learn more about combustion of organic compounds here: brainly.com/question/13272422
