Br₂ (l) + 2 NaI (s) → 2 NaBr (s) + I₂ (s)
Explanation:
Reacting bromide (Br₂) with sodium iodine (NaI) will produce sodium bromide (NaBr) and iodine (I₂).
To balance the equation the number of atoms of each element entering the reaction have to be equal to the number of atoms of each element leaving the reaction, in order to conserve the mass.
Br₂ (l) + 2 NaI (s) → 2 NaBr (s) + I₂ (s)
where:
l - liquid
s - solid
This is a single replacement reaction because an element in a compound is replaced by another element. Generally a single replacement reaction is represented as: A + BC → AC + B
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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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A technician mixes 80 ml of a 5% solution with 10 ml of water. the final percentage strength of the solution prepared is 40 %.
given that :
8 ml of a 5 % solution mix with 10 ml . that means the 80 mL of 5 % solution is diluted with water of 10 mL
therefore, 80 × 5 = 10 × x %
x % = 40 %
Therefore, the final percentage strength of the solution is 40 %
Thus, A technician mixes 80 ml of a 5% solution with 10 ml of water. the final percentage strength of the solution prepared is 40 %.
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Answer:
The Relative Formula Mass of Fe(NO₃)₂ is 179.8524 grams
Explanation:
The Relative Formula Mass is the mass of one mole of a compound expressed in grams, obtained by adding together the Relative Atomic Masses, RAM, of the elements which makes the compound
The Relative Formula Mass of a compound is the same as its Relative Molecular Mass
The relative formula mass of Fe(NO₃)₂ is given as follows;
The relative atomic mass of Fe = 55.845 amu
The relative atomic mass of nitrogen, N = 14.0067 amu
The relative atomic mass of oxygen, O = 15.999 amu
Therefore, we have;
The formula mass of Fe(NO₃)₂ = (55.845 + 2×(14.0067 + 3×15.999)) amu = 179.8524 amu
The Relative Formula Mass of Fe(NO₃)₂ = 179.8524 grams.