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:
N2(g) + 3H2(g) → 2 NH3(g)
Explanation:
N2(g) + H2(g) → NH3(g)
We start equaling the number of N atoms in both sides multiplying by 2 the NH3.
N2(g) + H2(g) → 2 NH3(g)
So we equals the H atoms (there are six in products sites)
N2(g) + 3 H2(g) → 2 NH3(g)
Answer:
The molarity of the solution is 0,31 M
Explanation:
We calculate the weight of 1 mol of NaCl from the atomic weights of each element of the periodic table. Then, we calculate the molarity, which is a concentration measure that indicates the moles of solute (in this case NaCl) in 1000ml of solution (1 liter)
Weight 1 mol NaCl= Weight Na + Weight Cl= 23 g + 35, 5 g= 58, 5 g
58, 5 g-----1 mol NaCl
13,1 g ---------x= (13,1 g x 1 mol NaCl)/58, 5 g= 0, 224 mol NaCl
727 ml solution------ 0, 224 mol NaCl
1000ml solution------x= (1000ml solutionx0, 224 mol NaCl)/727 ml solution
x=0,308 mol NaCl---> <em>The solution is 0,31 molar (0,31 M)</em>
The segment that represents melting is time (minutes) and temperature.
