The specific heat : c = 0.306 J/g K
<h3>Further explanation</h3>
Given
Heat = 35.2 J
Mass = 16 g
Temperature difference : 7.2 K =
Required
The specific heat
Solution
Heat can be calculated using the formula:
Q = mc∆T
Q = heat, J
m = mass, g
c = specific heat, joules / g ° C
∆T = temperature difference, ° C / K
Input the value :
c = Q / m.∆T
c = 35.2 / 16 x 7.2
c = 0.306 J/g K
Answer:
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Answer:
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Answer:
mass 1.25 Liters NH₃(gas) = 0.949 grams (3 sig-figs)
Explanation:
At STP (Standard Temperature-Pressure conditions => 0°C(=273K) and 1atm pressure, 1 mole <u>any</u> gas will occupy 22.4 Liters.
So, given 1.25 Liters ammonia gas at STP, convert to moles then multiply by formula wt. (17g/mole gives mass of NH₃.
moles NH₃(gas) = 1.25L NH₃(gas)/22.4L NH₄(gas)· NH₃(gas)mole⁻¹ = 0.0558 mole NH₃(gas).
Converting to grams NH₃(gas) = 0.0558 mole NH₃(gas) x 17 g·mol⁻¹ = 0.949 grams NH₃(gas).
