A rectangular block floats in pure water with 0.5 inch above the surface and 1.5 inches below the surface. When placed in an aqu
1 answer:
Answer:
- the specific gravity of the block is 0.75
- the specific gravity of the solution is 1.5
Explanation:
Given the data in the question;
first we find the specific gravity of a block SGB
SGB = ( block vol below / total block vol ) × the specific gravity of water
we substitute
SG = ( 1.5 / (1.5 + 0.5 ) ) × 1
SG = ( 1.5 / (1.5 + 0.5 ) ) × 1
SG = (1.5 / 2) × 1
SG = 0.75
Therefore, the specific gravity of the block is 0.75
specific gravity of solution SG
SG = (total block vol / block below ) × SG
we substitute
SG = ( 2 / 1 ) × 0.75
SG = 2 × 0.75
SG = 1.5
Therefore, the specific gravity of the solution is 1.5
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Answer: 25.13 g of and 20.56 g of
will be produced from 45.70 g of
Explanation:
To calculate the moles :
The balanced chemical equation is:
According to stoichiometry :
1 mole of produce = 1 mole of
Thus 0.571 moles of will require=
of
Mass of
1 mole of produce = 2 moles of
Thus 0.571 moles of will require=
of
Mass of
Thus 25.13 g of and 20.56 g of
will be produced from 45.70 g of