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
Scientific would be the word to fill in the blank
Answer:
C
Explanation:
75 mile shallow flat area just off coastlines
It’s ionic bond based on electrons gain/loss.
<span>B)<span>C2H6O<span>2
</span></span></span>
First, convert each percentage to grams: 38.7g, 9.70g, and 51.6g.
Next, calculate the number of moles of each element, based on the number of grams given.
C = 3.23 mol
H = 8.91 mol
O = 3.23 mol
Set up the ratio of moles of each element:
C3.34H9.70O3.23. Convert the decimals to whole numbers by dividing by the smallest subscript, 3.23.
The empirical formula is CH3O.
Now, compute the formula mass, which is 31. Finally, divide the molecular mass by the formula mass, 62/31 = 2. Multiple the subscripts by 2 to get the molecular formula.
