Answer:
ºC
Explanation:
We have to start with the variables of the problem:
Mass of water = 60 g
Mass of gold = 13.5 g
Initial temperature of water= 19 ºC
Final temperature of water= 20 ºC
<u>Initial temperature of gold= Unknow</u>
Final temperature of gold= 20 ºC
Specific heat of gold = 0.13J/gºC
Specific heat of water = 4.186 J/g°C
Now if we remember the <u>heat equation</u>:
We can relate these equations if we take into account that <u>all heat of gold is transfer to the water</u>, so:
Now we can <u>put the values into the equation</u>:
Now we can <u>solve for the initial temperature of gold</u>, so:
ºC
I hope it helps!
1 ba+2 br——>1 babr2
u just have to make sure u have the same number of each type of atom on either side of the equation:)
Missing question: What is the rate constant for the reaction?
<span>[RS2](mol L-1) Rate (mol/(L·s))
0.150 0.0394
0.250 0.109
0.350 0.214
0.500 0.438</span>
Chemical reaction: 3RS₂ → 3R + 6S.
Compare second and fourth experiment, when concentration is doubled, rate of concentration is increaced by four. So rate is:
rate = k·[RS₂]².
k = 0,438 ÷ (0,500)².
k = 1,75 L/mol·s.
Answer: The beginning stage you decide to depict the area, or position, of an object.
Explanation: starting point or position
Answer:
D. 5 moles
Explanation:
C3H8 + 5O2 → 4H2O + 3CO2
5 mol 3 mol
So, to make 3 mol CO2 5 mol O2 are needed.
