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Ray Of Light [21]
3 years ago
8

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

energy when burned. To illustrate this difficulty, calculate the number of liters of water that must be expended to absorb the energy released by burning 1.00 L 1.00 L of crude oil, if the water has its temperature raised from 23.5 °C 23.5 °C to 100 °C 100 °C , it boils, and the resulting steam is raised to 315 °C.
Chemistry
1 answer:
Alborosie3 years ago
4 0

Answer:

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

c=4186J/kg°C

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

b) since the density of crude is less than the density of water,and 80L of additional water is added, it'll make the crude to float on water thus inhibiting the extinguishing process

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Answer:

mole fraction of NaCl = 0.03145.

mole fraction of water = 0.9686.

Explanation:

  • Mole fraction is an expression of the concentration of a solution or mixture.
  • It is equal to the moles of one component divided by the total moles in the solution or mixture.
  • The summation of mole fraction of all mixture components = 1.

mole fraction of NaCl = (no. of moles of NaCl) / (total no. of moles).

<em>no. of moles of NaCl = mass/molar mass </em>= (6.87 g)/(58.44 g/mol) = 0.1176 mol.

<em>no. of moles of water = mass/molar mass</em> = (65.2 g)/(18.0 g/mol) = <em>3.622 mol.</em>

<em></em>

∴ mole fraction of NaCl = (no. of moles of NaCl) / (total no. of moles) = (0.1176 mol)/(0.1176 mol + 3.622 mol) = 0.03145.

<em>∵ mole fraction of NaCl + mole fraction of water = 1.0.</em>

∴ mole fraction of water = 1.0 - mole fraction of NaCl = 1.0 - 0.03145 = 0.9686.

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3 years ago
Use the equation 2cu+o2=2cuo to find how many miles of copper react to form 3.64 mil cuo
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<h3><u>Answer;</u></h3>

3.64 moles

<h3><u>Explanation;</u></h3>

From the equation;

2Cu + O2 → 2CuO

Mole ratio of Cu : CuO = 2: 2 equivalent to 1: 1

Moles of CuO = 3.64 Moles

Therefore;

moles of Cu will be ;

3.64 × 1 = 3.64 moles

Moles of Cu = <u>3.64 moles </u>

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A breakfast of cereal, orange juice, and milk might contain 230 nutritional Calories. How much is this in Joules?
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Bauxite must go through two processes to produce aluminum metal. The yield of the Bayer process, which extracts aluminum oxide f
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The yield of aluminium obtained from 1 m^3 of bauxite is 419810 g

<h3>What is a Percent yield</h3>

A percent yield of a substance measures the amount of the substance actually obtained as a percentage ratio of expected yield.

Percent yield = actual yield / expected yield × 100%

<h3>How to calculate the mass of aluminium obtained from bauxite </h3>

From the data given:

40 % of the bauxite is converted to aluminium oxide.

Volume of bauxite = 1 m^3

40 % of 1 m^3 = 0.4 m^3

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density of aluminium oxide = 3965 kg/m^3

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mass of aluminium oxide = 0.4 × 3965 kg

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molar mass of aluminium oxide = 102 g

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Percentage mass of aluminium in aluminium oxide = 54/102 × 100

Percentage mass of aluminium in aluminium oxide = 52.94 %

Expected mass of aluminium from aluminium oxide = 52.94 × 1586

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Actual yield of aluminium = 419.81 kg

mass of aluminium in grams = 419810 g

Therefore, mass of aluminium obtained from 1 m^3 of bauxite is 419810 g

Learn more about percent yield at: brainly.com/question/8638404

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