Atomic mass Cu = 63.546 a.m.u
63.546 g ---------------- 6.02x10²³ atoms
22 g --------------------- ??
22 x (6.02x10²³ ) / 63.546 => 2.08x10²³ atoms
hope this helps!
Answer:
32.6 %
Explanation:
Given data
- Mass of sucrose (solute): 22.8 grams
- Mass of water (solvent): 47.1 grams
Step 1: Calculate the mass of the solution
The mass of the solution is equal to the sum of the mass of the solute and the mass of the solvent.
m(solution) = m(solute) + m(solvent)
m(solution) = 22.8 g + 47.1 g
m(solution) = 69.9 g
Step 2: Calculate the percent-by-mass of sucrose in the solution
We will use the following expression.

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
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