Answer:
The total pressure of the mixture in the tank of volume 6.25 litres at 51°C is 1291.85 kPa.
Explanation:
For N2,
Pressure(P₁)=125 kPa
Volume(V₁)=15·1 L
Temperature (T₁)=25°C=25+273 K=298 K
Similarly, for Oxygen,
Pressure(P₂)= 125 kPa
Volume(V₂)= 44.3 L
Temperature(T₂)=25°C= 298 K
Then, for the mixture,
Volumeof the mixture( V)= 6.25 L
Pressure(P)=?
Temperature (T)= 51°C = 51+273 K=324 K
Then, By Combined gas laws,
or, 
or, 
or, 
∴P=1291.85 kPa
So the total pressure of the mixture in the tank of volume 6.25 litres at 51°C is 1291.85 kPa.
Answer:
0.74 grams of methane
Explanation:
The balanced equation of the combustion reaction of methane with oxygen is:
it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.
firstly, we need to calculate the number of moles of both
for CH₄:
number of moles = mass / molar mass = (3.00 g) / (16.00 g/mol) = 0.1875 mol.
for O₂:
number of moles = mass / molar mass = (9.00 g) / (32.00 g/mol) = 0.2812 mol.
- it is clear that O₂ is the limiting reactant and methane will leftover.
using cross multiplication
1 mol of CH₄ needs → 2 mol of O₂
??? mol of CH₄ needs → 0.2812 mol of O₂
∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol
so 0.14 mol will react and the remaining CH₄
mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol
now we convert moles into grams
mass of CH₄ left over = no. of mol of CH₄ left over * molar mass
= 0.0469 mol * 16 g/mol = 0.7504 g
So, the right choice is 0.74 grams of methane
Answer:
6960 J/kg°C
Explanation:
specific heat= mass×specific heat capacity×increase in temperature
specific heat= 0.240×1450×20= 6960 J/kg°C
hope it helps!
Answer: b) Crash 2; the force on the cart was stronger in this crash, so the force on the skateboard was also stronger.
Explanation:
Answer:
0.0000098 should be the answer
Explanation:
