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%
The mass in grams of nitric acid that is required to react with 454g C7H8 is 932.72 grams
calculation
find the moles of C7H8 used
moles = mass/molar mass
= 454 g/92 = 4.935 moles
balanced reacting equation
C7H8 +3 HNO3 = C7H5N3O6 +3 H20
by use of mole ratio between C7H8 to HNo3 which is 1:3 the moles of HNO3 =4.935 x3 = 14.805 moles
mass of HNo3 = moles x molar mass
= 14.805 x 63 = 932.72 grams
part B
the mass of C7H5N3o6 = 2045.5 grams
calculate the moles of C7H8
= 829 g/92 g/mol = 9.011 moles
by use of mole ratio between C7H8 to C7H5N3O6 which is 1:1 the moles of C7H5N3O6 is also = 9.011 moles
mass of C7H5N3O6 is therefore = moles x molar mass
=9.011 x227 = 2045.5 grams
Answer:
There are three ways that scientists have proved that these sub-atomic particles exist. They are direct observation, indirect observation or inferred presence and predictions from theory or conjecture. Scientists in the 1800's were able to infer a lot about the sub-atomic world from chemistry.
Explanation:
Hope this helps
If you mean what group of elements react the most, the answer is the alkali metals and the halogens because they both only either need to gain or lose one electron. If you mean the most reactive element, it would be fluorine because it has the most electronegativity.
