To
determine the empirical formula of the compound given, we need to determine the ratio of each element in the compound. To do that we assume to have 100 grams sample
of the compound with the given composition. Then, we calculate for the number
of moles of each element. We do as follows:<span>
mass moles
C 56.79 4.73
H 6.56 6.50
O 28.37 1.77
N 8.28 0.59
Dividing the number of moles of each element with
the smallest value, we will have the empirical formula:
</span> moles ratio
C 4.73 / 0.59 8
H 6.50 / 0.59 11
O 1.77 / 0.59 3
N 0.59 / 0.59 1<span>
</span><span>
The empirical formula would be C8H11O3N.</span>
Answer:
2AgNO3 + Cu = Cu(NO3)2 + 2Ag
Explanation:
First you see which side has the most elements. Which is Cu(NO3)2 + Ag. But, both sides have the same elements? But, on the reactants side, there is 2 of NO3. On the products side there is only one.
Reactants: Products:
Cu = 1 Cu = 1
NO3 = 1 NO3 = 2
Ag = 1 Ag = 1
They are all equal, except for NO3. So on the reactants side, you add a two to make it even.
2AgNO3 + Cu = Cu(NO3)2 + Ag
Reactants: Products:
Cu = 1 Cu = 1
NO3 = 2 NO3 = 2
Ag = 2 Ag = 1
Now, the NO3 is equal, But the Ag isn't. But, you can add a 2 on the <u>products</u> side so the whole equation becomes equal.
2AgNO3 + Cu = Cu(NO3)2 + 2Ag
Reactants: Products:
Cu = 1 Cu = 1
NO3 = 2 NO3 = 2
Ag = 2 Ag = 2
Answer:
their are a it is in the chemicle
Explanation:18 carbon and 4 hydrogen
so it is a toatle of 22 atoms
Answer:
The answer to your question is 8.75 x 10 ²⁵ atoms of Magnesium
Explanation:
Data
Atomic mass of Magnesium = 24.31 g
Atoms of magnesium = 3.60 x 10²⁴
Process
Solve this porblem using a rule of three
1 atom of Magnesium --------------- 24.31 g
3.60 x 10²⁴ ---------------- x
x = (3.60 x 10²⁴ x 24.31) / 1
x = 8.75 x 10 ²⁵ atoms
