The chemical reaction would be expressed as follows:
HBr + LiOH = LiBr + H2O
We are given the volumes and corresponding concentration to be used for the reaction. We use these values to solve for the concentration of the other reactant. We do as follows:
0.253 mol LiOH / L solution ( 0.01673 L ) ( 1 mol HBr / 1 mol LiOH ) = 0.00423 HBr needed
Concentration of HBr =0.00423mol / .010 L = 0.423 M HBr
The answer is (4) at the cathode, where reduction occurs. The Na+ gains one electron and become Na(l). So the reaction occurs at cathode and is reduction reaction.
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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Answer:
In the last 25 years, the fish hunt in the Chesapeake Bay occurred very fastly due to which the fish industry enhanced. However, this rapid fish industrialization caused many of the fish species to become endangered. Hence, the fish industries started to use two basic management techniques which were conservation and allocation.
