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:
First balance the equation
C2H6 + 7/2O2 2CO2 +3H2O
for one mole of C2H6 there are 7/2 mole of O2 required. so for4.50 moles you require 4.50 x 7/2 = 15.75 moles of O2.
The answer is either a superphosphate or a triple superphosphate
<u>Answer:</u> The mass of phosphorus that is present for given amount of calcium is 28.53 g.
<u>Explanation:</u>
We are given:
Mass of calcium = 50 grams
The chemical formula of calcium phosphate is
Molar mass of calcium = 40 g/mol
Molar mass of phosphorus = 31 g/mol
In 1 mole of calcium phosphate, 120 grams of calcium is combining with 62 grams of phosphorus.
So, 50 grams of calcium will combine with = of phosphorus.
Hence, the mass of phosphorus that is present for given amount of calcium is 28.53 g.