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asambeis [7]
3 years ago
8

A sample of phosphonitrilic chloride, PNCl2, contains 0.170 mol of the compound. What is the mass of the sample, in grams?

Chemistry
1 answer:
Galina-37 [17]3 years ago
5 0

Answer: 19.7 grams PNCl2

Explanation: To find the mass of the sample in grams, all you have to do is convert. You need to multiply 0.170 mol by the unit multiplier 115.89 grams PNC12 (molecular mass - rounded) / 1 mol PNC12. This will cancel out the moles of PNCl2, leaving you with grams as the units. Now, all you have to do is multiply. 0.170*115.89 = 19.7 grams PNCl2 (rounded, again).

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It is difficult to extinguish a fire on a crude oil tanker, because each liter of crude oil releases 2.80 × 10 7 J 2.80×107 J of
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The question is incorrect and incomplete. Here's the correct question:

It is difficult to extinguish a fire on a crude oil tanker, because each liter of crude oil releases 2.80 × 10 7 J  of energy when burned. To illustrate this difficulty,a) calculate the number of liters of water that must be expended to absorb the energy released by burning 1.00 L  of crude oil, if the water has its temperature raised from 23.5 °C to 100 °C , it boils, and the resulting steam is raised to 315 °C. b)Discuss additional complications caused by the fact that crude oil has less density than water.

Explanation:

Q= mc ΔT

Q= heat energy

m is mass

ΔT is change in temperature and c is specific heat capacity

calculating heat for latent heat of vaporisation

Q= ml where l is latent heat of vaporisation

a) Total heat energy used= heat required to raise temperature from 23.5 °C to 100 °C, heat required to boil water and heat required to further raise temperature from 100 °C to 315°C

Q = mc ΔT₁ + mL + mc ΔT₂

Q = m(c ΔT₁ + L + c ΔT₂)

m= Q÷(c ΔT₁ + L + c ΔT₂)

Q= 2.8 X 10⁷ J

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L=2256 x 10³J/kg

ΔT₁=76.5°C(100°C-23.5°C)

ΔT₂= 215°C(315°C-100°C)

(c ΔT₁ + L + c ΔT₂)= 4186J/kg°C *76.5°C + 2256 x 10³J/kg + 4186J/kg°C*215°C =3476219J/Kg

m= 2.8 x 10⁷J ÷3476219J/Kg

m =80.54 Kg

volume = mass÷ density

=80.54kg ÷ 10³kg/m³( density of water)

=0.0854m³

0.001m³ = 1 lL0.08054m³= 0.08054m³ /0.001m³= 80.54L

VOLUME is 80.54litres

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