One is correct, <span>carbon dioxide, methane, and water vapor are the three major greenhouse gases found in the atmosphere.</span>
Answer:
Vapor pressure of solution is 78.2 Torr
Explanation:
This is solved by vapor pressure lowering:
ΔP = P° . Xm . i
Vapor pressure of pure solvent (P°) - vapor pressure of solution = P° . Xm . i
NaCl → Na⁺ + Cl⁻ i = 2
Let's determine the Xm (mole fraction) These are the moles of solute / total moles.
Total moles = moles of solvent + moles of solute
Total moles = 0.897 mol + 0.182 mol → 1.079 mol
0.182 / 1.079 = 0.168
Now we replace on the main formula:
118.1° Torr - P' = 118.1° Torr . 0.168 . 2
P' = - (118.1° Torr . 0.168 . 2 - 118.1 Torr)
P' = 78.2 Torr
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
Answer: Okay so here's the order lol from top to bottom
2, 1, 3, 4, 5
Explanation:
