Answer:
Work done in Joules = -14865.432J
ΔU = -262.94KJ
Explanation:
The steps are as shown in the attachment
Answer:
Explanation:
we know that specific heat is the amount of heat required to raise the temperature of substance by one degree mathmeticaly
Q=mcΔT
ΔT=T2-T1
ΔT=26.8-10.2=16.6
C for water is 4.184
therefore
Q=1.00*4.184*16.6
Q=69.4 j
now we have to covert joule into calorie
1 calorie =4.2 j
x calorie=69.4 j/2
so 69.4 j =34.7 calorie thats why 34.7 calorie heat is required to raise the temperature of water from 10.2 to 26.8 degree celsius
Answer:

Explanation:
Mole ratio


Moles of CO₂

Use formula to find mass

Relative molecular mass of CO₂ = 12 + 16 × 2 = 44


Answer:
390 m
Explanation:
Martha ran 420 m
Cole ran 60 m less than Martha = 420 m - 60 m = 360 m
The distance Kate ran is right in the middle of Martha and Cole's distances
Kate distance = (420 + 360) / 2 = 780 / 2 = 390 m
Question:
At standard temperature and pressure, the volume of a tire is 3.5L. What is the new pressure if the temperature outside is 296k and its weight causes the volume of the gas is 2.0 L?
Answer:
The new pressure is: 1.896 atm
Explanation:
At standard temperature and pressure, we have:



Outside, we have:


Required
Determine the new pressure
Using combined gas law, we have:

This gives:

Solve for 


