Answer : The amount of heat evolved by a reaction is, 4.81 kJ
Explanation :
Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the water
![q=[q_1+q_2]](https://tex.z-dn.net/?f=q%3D%5Bq_1%2Bq_2%5D)
![q=[c_1\times \Delta T+m_2\times c_2\times \Delta T]](https://tex.z-dn.net/?f=q%3D%5Bc_1%5Ctimes%20%5CDelta%20T%2Bm_2%5Ctimes%20c_2%5Ctimes%20%5CDelta%20T%5D)
where,
q = heat released by the reaction
= heat absorbed by the calorimeter
= heat absorbed by the water
= specific heat of calorimeter = 
= specific heat of water = 
= mass of water = 254 g
= change in temperature = 
Now put all the given values in the above formula, we get:
![q=[(783J/^oC\times -2.28^oC)+(254g\times 4.184J/g^oC\times -2.28^oC)]](https://tex.z-dn.net/?f=q%3D%5B%28783J%2F%5EoC%5Ctimes%20-2.28%5EoC%29%2B%28254g%5Ctimes%204.184J%2Fg%5EoC%5Ctimes%20-2.28%5EoC%29%5D)
Therefore, the amount of heat evolved by a reaction is, 4.81 kJ
<span>The half-life of 9 months is 0.75 years.
2.0 years is 2.0/0.75 = 2.67 half-lives.
Each half-life represents a reduction in the amount remaining by a factor of two, so:
A(t)/A(0) = 2^(-t/h)
where A(t) = amount at time t
h = half-life in some unit
t = elapsed time in the same unit
A(t)/A(0) = 2^(-2.67) = 0.157
15.7% of the original amount will remain after 2.0 years.
This is pretty easy one to solve. I was happy doing it.</span>
Answer:
51.66 mol H2O (steam)
Explanation:
5.74 mol C3H18 x 18 mol H2O/ 2 mol C3H18 = 51.66 mol H2O (steam)
Answer:
We can use heat = mcΔT to determine the amount of heat, but first we need to determine ΔT. Because the final temperature of the water is 55°C and the initial temperature is 20.0°C, ΔT is as follows:
ΔT = Tfinal − Tinitial = 55.0°C − 20.0°C = 35.0°C
given the specific heat of water as 1 cal/g·°C. Substitute the known values into heat = mcΔT and solve for amount of heat:
= heat=(75.0 g)(1 cal/ g· °C )(35.0°C) =
= 75x1x35=2625 cal
Answer:
The answer to your question is:
Explanation:
Reaction
SnCl₂ + 2KMnO₄ ⇒ 2 KCl + Sn(MnO₄)₂
1 ---- Sn ---- 1
2 ---- K ----- 2
2 ---- Mn ---- 2
8 ---- O ---- 8
2 ---- Cl ---- 2
