Empirical formula is the simplest formula showing the simplest ratio of atoms in a compound. Calculated as shown;
we start by calculating the number of moles of each atom;
moles of nickel = 9.11 g ÷ 58.7 g = 0.155 moles
moles of fluorine = 5.89 g ÷ 19 g = 0.31 moles
Then we get the ratio of the moles of nickel to that of flourine
That is 0.155 : 0.31 (dividing by the smallest)
0.155/0.155 : 0.31/0.155
we get 1:2 ( the simplest ratio)
Therefore the empirical formula is nif2
Answer: 3.59
Explanation:
(2.06)(1.743)(1.00)
2.06 × 1.743 × 1.00
= 3.59058
Two of the multiplied digits are represented in 3 significant figures. Therefore, for correct representation, the result of the product should be written to three significant figures.
3.59058 to 3 significant figures:
First three digits = 3.59
Fourth digit '0' is less than 5, and thus rounded to 0 with other succeeding digits
Therefore, (2.06)(1.743)(1.00) to 3 significant figures equals :
3.59
Answer:
Let the mixture is X% by mass of CuSO
4
.5H
2
O and 100 - X % by mass of MgSO
4
.7H
2
O. 5.0 g of mixture will contain 0.05X g CuSO
4
.5H
2
O and 5.0 - 0.05X g MgSO
4
.7H
2
O
The molar masses of CuSO
4
.5H
2
O and MgSO
4
.7H
2
O are 249.7 g/mol and 246.5 g/mol respectively.
The number of moles of CuSO
4
.5H
2
O=
249.7
0.05X
=2.00×10
−4
X moles.
Explanation:
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