Answer: 0.5 moles
Explanation:
Cr2(SO3)2 is the chemical formula for chromium sulphate.
Given that,
Amount of moles of Cr2(SO3)2 (n) = ?
Mass of Cr2(SO3)2 in grams = 128.9g
For molar mass of Cr2(SO3)2, use the atomic masses:
Chromium, Cr = 52g;
Sulphur, S = 32g;
Oxygen, O = 16g
Cr2(SO3)2 =
(52g x 2) + [(32g + 16g x 3) x 2]
= 104g + [(32g + 48g) x 2]
= 104g + [80g x 2]
= 104g + 160g
= 264g/mol
Since, n = mass in grams / molar mass
n = 128.9g / 264g/mol
n = 0.488 mole [Round the value of n to the nearest tenth which is 0.5
Thus, there are 0.5 moles in 128.9 grams of Cr2(SO3)2
Answer:
5 L.
Explanation:
From the question given above, the following data were obtained:
Initial volume (V1) = 10 L
Initial pressure (P1) = 2.5 atm
Final pressure (P2) = 5 atm
Final volume (V2) =.?
Since the temperature is constant, we shall apply the Boyle's law equation to determine the new volume of the gas. This can be obtained as follow:
P1V1 = P2V2
2.5 × 10 = 5 × V2
25 = 5 × V2
Divide both side by 5
V2 =25/5
V2 = 5 L
Thus, the new volume of the gas is 5 L
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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