A very long tube with a cross-sectional area of 1.00 cm2 is filled with mercury to a height of 76.0 cm. At what height would water stand in this tube if it were filled with a mass of water equal to that of the mercury? (The density of mercury is 13.60 g/cm3 and the density of water is 1.00 g/cm3.)

Answers

Answer 1

The water would stand at a height of 1033.6 cm

If the mass of the mercury, m equals the mass of the water, m' at its height,h', then the weight of the mercury W equals the weight of the water, W'

So, W = W'

mg = m'g

ρV = ρ'V' where ρ = density of mercury = 13.60 g/cm³, V = volume of mercury = Ah where A = cross-sectional area of tube = 1.00 cm and h = height of mercury = 76.0 cm, ρ' = density of water = 1.00 g/cm³, V = volume of water = Ah' where A = cross-sectional area of tube = 1.00 cm and h' = height of water.

So, ρV = ρ'V'

ρAh = ρ'Ah'

h' = ρAh/ρ'A

h' = ρh/ρ'

Substituting the values of the variables into the equation, we have

h' = ρh/ρ'

h' = 13.60 g/cm³ × 76.0 cm/1.00 g/cm³

h' = 13.60 × 76.0 cm

h' = 1033.6 cm

So, the water would stand at a height of 1033.6 cm

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can u tell the question in english

Each of the following sets of quantum numbers is supposed
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Answers

Answer:

= 0.198 mole H₂S in excess

Explanation:

A quick way to determine limiting reactant is to convert reactant values given to moles and divide by respective coefficient. The smaller value is the limiting reactant other reactants will be in excess. However, when working problem, use mole values given to solve, not results of division.

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Tingnan ang pinakamahuhusay na kagawian at patnubay sa pampublikong kalusugan para mapanatiling ligtas ang iyong mga customer at tauhan.

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Answers

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Answers

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Answer:

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