The empirical formula : C₁₂H₄F₇
The molecular formula : C₂₄H₈F₁₄
<h3>Further explanation</h3>
mol C (MW=12 g/mol)
mol H(MW=1 g/mol) :
mol F(MW=19 g/mol)
mol ratio of C : H : O =1.52 : 0.51 : 0.89=3 : 1 : 1.75=12 : 4 : 7
Empirical formula : C₁₂H₄F₇
(Empirical formula)n=molecular formula
( C₁₂H₄F₇)n=562 g/mol
(12.12+4.1+7.19)n=562
(281)n=562⇒ n =2
Molecular formula : C₂₄H₈F₁₄
Answer:
2H⁺(aq) + 2OH⁻(aq) --> 2H2O(l)
Explanation:
2HBr(aq)+Ba(OH)2(aq)⟶2H2O(l)+BaBr2(aq)
We break the compounds into ions. Only compounds in the aqueous form can be turned into ions.
The ionic equation is given as;
2H⁺(aq) + 2Br⁻(aq) + Ba²⁺(aq) + 2OH⁻(aq) --> 2H2O(l) + Ba²⁺(aq) + 2Br⁻(aq)
Upon eliminating the spectator ions; The net equation is given as;
2H⁺(aq) + 2OH⁻(aq) --> 2H2O(l)
Answer:
The average atomic weight = 121.7598 amu
Explanation:
The average atomic weight of natural occurring antimony can be calculated as follows :
To calculate the average atomic mass the percentage abundance must be converted to decimal.
121 Sb has a percentage abundance of 57.21%, the decimal format will be
57.21/100 = 0.5721 . The value is the fractional abundance of 121 Sb .
123 Sb has a percentage abundance of 42.79%, the decimal format will be
42.79/100 = 0.4279. The value is the fractional abundance of 123 Sb .
Next step is multiplying the fractional abundance to it masses
121 Sb = 0.5721 × 120.904 = 69.169178400
123 Sb = 0.4279 × 122.904 = 52.590621600
The final step is adding the value to get the average atomic weight.
69.169178400 + 52.590621600 = 121.7598 amu
Answer:
a motor that requires no energy input once it is running
<u>Answer:</u> The speed of light in miles per minutes is 
<u>Explanation:</u>
We are given the speed of light is 
and we need to convert it into miles/min. So, we use the converion factor:
1 hour = 60 minutes
Converting that quantity into miles/minutes, we get:
Hence, the speed of light in miles per minutes is
