Answer:
D)subscript of C in molecular formula = n x subscript of C in empirical formula
Explanation:
THIS IS THE COMPLETE QUESTION BELOW;
.The empirical formula for a compound is CH2. If n is a whole number, which shows a correct relationship between the molecular formula and the empirical formula? a)<br /><br /> empirical formula mass / molecular mass = n<br /><br /> B) molecular mass = element mass / empirical formula mass ´ 100<br /><br /> c) subscript of H in empirical formula = 2  subscript of H in molecular formula<br /><br /> D) subscript of C in molecular formula = n  subscript of C in empirical formula<br /><br />
An empirical formula can be regarded as "shorten form" of a molecular formula. Instance of this is
A compounds CH4, C2H8, C4H12... with empirical formula of CH4. In this case a constant "n" represent the difference that exist between empirical formula and molecular formula, "n" which is a whole number, molecular formula is the numerator.
Therefore, subscript of C in molecular formula = n x subscript of C in empirical formula
0 degrees Celsius, 32 degrees Fahrenheit. So the answer is a
Answer: When atoms combine through chemical bonding, they form compounds—unique structures composed of two or more atoms. The basic composition of a compound can be indicated using a chemical formula.
Hope I was able to help! Mark me brainly it would help a lot!:)
Answer:
The given element is Radon because its atomic weight is 222 amu.
Explanation:
Given data:
Percentage of A-219 = 13.92%
Percentage of B-222 = 72.16%
Percentage of C-225 = 13.92%
Atomic weight of element = ?
Solution:
Average atomic mass = (abundance of A isotope × its atomic mass) +(abundance of B isotope × its atomic mass) + (abundance of C isotope × its atomic mass) / 100
Average atomic mass = (13.92×219)+(72.16×222) + (13.92×225)/100
Average atomic mass = 3048.48 + 16019.52 +3132/ 100
Average atomic mass = 22200 / 100
Average atomic mass = 222 amu.
The given element is Radon because its atomic weight is 222 amu.
