Answer:
Specific heat of water = 33.89 KJ
Explanation:
Given:
mass of water = 81 gram
Initial temperature = 0°C
Final temperature = 100°C
Specific heat of water = 4.184
Find:
Required heat Q
Computation:
Q = Mass x Specific heat of water x (Final temperature - Initial temperature)
Q = (81)(4.184)(100-0)
Q = 33,890.4
Specific heat of water = 33.89 KJ
One thing to notice in the question is, we are asked about molecular oxygen that has formula O2 not atomic oxygen O.
As we are asked about molecular oxygen, we will answer the question in terms of number of molecules that are present in 16 grams of molecular oxygen.
To get the number of molecules present in 16 grams of O2, we will use the formula:
No. of molecules = no. of moles x Avogadro's number (NA)----- eq 1)
As we know:
The number of moles = mass/ molar mass of molecule
Here we have been given mass already, 16 grams and the molar mass of O2 is 32 grams.
Putting the values in above formula:
= 16/32
= 0.5 moles
Putting the number of moles and Avogadro's number (6.02 * 10^23) in eq 1
No. of molecules = 0.5 x 6.02 * 10^23
=3.01 x 10^23 molecules
or 301,000,000,000,000,000,000,000 molecules
This means that 16 grams of 3.01 x 10^23 molecules of oxygen.
Hope it helps!
I think it is B because you didn’t type the options
Answer:
648.5 mL
Explanation:
Here we will assume that the pressure of the gas is constant, since it is not given or specified.
Therefore, we can use Charle's law, which states that:
"For an ideal gas kept at constant pressure, the volume of the gas is proportional to its absolute temperature"
Mathematically:
where
V is the volume of the gas
T is its absolute temperature
The equation can be rewritten as
where in this problem we have:
is the initial volume of the gas
is the initial temperature
is the final temperature
Solving for V2, we find the final volume of the gas:
