Answer:
0.924 g
Explanation:
The following data were obtained from the question:
Volume of CO2 at RTP = 0.50 dm³
Mass of CO2 =?
Next, we shall determine the number of mole of CO2 that occupied 0.50 dm³ at RTP (room temperature and pressure). This can be obtained as follow:
1 mole of gas = 24 dm³ at RTP
Thus,
1 mole of CO2 occupies 24 dm³ at RTP.
Therefore, Xmol of CO2 will occupy 0.50 dm³ at RTP i.e
Xmol of CO2 = 0.5 /24
Xmol of CO2 = 0.021 mole
Thus, 0.021 mole of CO2 occupied 0.5 dm³ at RTP.
Finally, we shall determine the mass of CO2 as follow:
Mole of CO2 = 0.021 mole
Molar mass of CO2 = 12 + (2×16) = 13 + 32 = 44 g/mol
Mass of CO2 =?
Mole = mass /Molar mass
0.021 = mass of CO2 /44
Cross multiply
Mass of CO2 = 0.021 × 44
Mass of CO2 = 0.924 g.
Answer : The correct option is, (D) 
Solution :
Formula used :
where,
Q = heat released = -1300 J
m = mass of water = 40 g
c = specific heat of water = 
= final temperature = ?
= initial temperature = 
Now put all the given values in the above formula, we get the final temperature of water.
Therefore, the final temperature of the water is,
I believe the answer is 8
Answer:
The required ratio would be 2:1
Answer:
its "a" nd "C" Big Dawg if im wrong im sorry dawg been a min since i took a test like dis
Explanation:
