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marysya [2.9K]
3 years ago
9

determine the specific heat of a mayterial if a 35g sample of the material absorbs 48J as it is heated from 298K to 313K

Chemistry
1 answer:
Oksi-84 [34.3K]3 years ago
6 0
Data:

m = 35g
Q = 48 j
T1 = 298K
T2 = 313K
Cs = ?

Fórmula: Q = m*Cs*(T2 - T1) => Cs = Q / [m (T2 - T1)]

Cs = 48 j / [35g (313K - 298K] = 0.0914 j / g*K = 91.4 j / kg*K
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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

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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

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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

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  • 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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