Answer:
c) (12×0.9889) + (13×0.01108)
Explanation:
Given data:
Percentage of C-12 = 98.89%
Percentage of C-13 = 1.108%
Atomic mass = ?
Solution:
98.89/100 = 0.9889
1.108/ 100 = 0.01108
Atomic mass = (12×0.9889) + (13×0.01108)
Atomic mass = (11.8668 + 0.144034)
Atomic mass = 12.01084
Answer:
Mass = 135.66 ×10⁻²¹ g
Explanation:
Given data:
Number of molecules of CuSO₄= 5.119×10²
Mass of CuSO₄= ?
Solution:
The given problem will solve by using Avogadro number.
1 mole contain 6.022×10²³ molecules
5.119×10² molecules ×1 mol / 6.022×10²³ molecules
0.85×10⁻²¹ mol
Mass in grams:
Mass = number of moles × molar mass
Mass = 0.85×10⁻²¹ mol × 159.6 g/mol
Mass = 135.66 ×10⁻²¹ g
First let us see what
kind of bonds are formed in the compound. By drawing the structure, we see that
the kind of bonds are:
N =- triple bond -= C –
O
<span>So there is only
single bond between C and O therefore the hybridization of C is sp.</span>
So to balance an equation, you need to get the same amount of each type of element on either side of the --> . So you pretty much are given the subscripts in the equations and you need to add coefficients (just normal numbers) in front of any formula that needs it, keeping anything balance.

turns into

These coefficient numbers are the molar ratios, so 2 moles of KCl3 for every 3 moles of O2 so 1. 3:2
Then you can use these ratios of find out how many moles of one thing are needed if you are given the amount of another.

and use cross multiplication to solve for whatever you don't know
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