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Slav-nsk [51]
4 years ago
11

What mass of lead (density 11.4 g/cm3) would have an identical volume to 28.5 g of mercury (density 13.6 g/cm3)?

Chemistry
1 answer:
kipiarov [429]4 years ago
6 0

The density of a material is its unique property which is different for different substances. The volume or mass of a compound can be determined from the other given mass of a compound, if the density of both the material is available. We know density of a material is equal to the mass per unit volume. From the given data 13.6 g of mercury has 1 cm³ volume. Thus 28.5g of mercury will have \frac{28.5}{13.6}=2.095 cm³ volume. Now as per the given density value of lead 11.4g of lead has 1 cm³ volume. Thus 2.095 cm³ volume of lead has (2.095×11.4)=23.889g.

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what is the limiting reactant when 1.50g of lithium and 1.50 g of nitrogen combine to form lithium nitride, a component of advan
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<h3>Answer:</h3>

Limiting reactant is Lithium

<h3>Explanation:</h3>

<u>We are given;</u>

  1. Mass of Lithium as 1.50 g
  2. Mass of nitrogen is 1.50 g

We are required to determine the rate limiting reagent.

  • First, we write the balanced equation for the reaction

6Li(s) + N₂(g) → 2Li₃N

From the equation, 6 moles of Lithium reacts with 1 mole of nitrogen.

  • Second, we determine moles of Lithium and nitrogen given.

Moles = Mass ÷ Molar mass

Moles of Lithium

Molar mass of Li = 6.941 g/mol

Moles of Li = 1.50 g ÷ 6.941 g/mol

                   = 0.216 moles

Moles of nitrogen gas

Molar mass of Nitrogen gas is 28.0 g/mol

Moles of nitrogen gas = 1.50 g ÷ 28.0 g/mol

                                     = 0.054 moles

  • According to the equation, 6 moles of Lithium reacts with 1 mole of nitrogen.
  • Therefore, 0.216 moles of lithium will require 0.036 moles (0.216 moles ÷6) of nitrogen gas.
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Thus, Lithium is the limiting reagent while nitrogen is in excess.

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