Answer:
0.733 J/g °C
Explanation:
<u>Step 1 : Write formule of specific heat</u>
<u />
Q
=
mc
Δ
T
with Q = heat transfer (J)
with m = mass of the substance
with c = specific heat ⇒ depends on material and phase ( J/g °C)
Δ
T = Change in temperature
<u />
in this case :
Q = 38.5 J
<u>m = 17.5g</u>
<u>c= has to be determined</u>
<u>Δ
T = 3 (°C)</u>
<u />
<u>Step 2: Calculating specific heat</u>
⇒ via the formule Q
=
mc
Δ
T
38.5 J = 17.5g * c * 3
38.5 = 52.5 *c
c = 38.5 / 52.5
c = 0.733 J/g °C ⇒ 0.733 is reported to three significant digits due to the precision.
The specific heat of this metal is 0.733 J/g °C
The question is incomplete, here is the complete question:
The rate of certain reaction is given by the following rate law:
![rate=k[H_2]^2[I_2]^2](https://tex.z-dn.net/?f=rate%3Dk%5BH_2%5D%5E2%5BI_2%5D%5E2)
At a certain concentration of 
and 
, the initial rate of reaction is 4.0 × 10⁴ M/s. What would the initial rate of the reaction be if the concentration of
Answer : The initial rate of the reaction will be, 
Explanation :
Rate law expression for the reaction:
As we are given that:
Initial rate = 4.0 × 10⁴ M/s
Expression for rate law for first observation:
![4.0\times 10^4=k[H_2]^2[I_2]^2](https://tex.z-dn.net/?f=4.0%5Ctimes%2010%5E4%3Dk%5BH_2%5D%5E2%5BI_2%5D%5E2)
....(1)
Expression for rate law for second observation:
![R=k(\frac{[H_2]}{2})^2[I_2]^2](https://tex.z-dn.net/?f=R%3Dk%28%5Cfrac%7B%5BH_2%5D%7D%7B2%7D%29%5E2%5BI_2%5D%5E2)
....(2)
Dividing 2 by 1, we get:
![\frac{R}{4.0\times 10^4}=\frac{k(\frac{[H_2]}{2})^2[I_2]^2}{k[H_2]^2[I_2]^2}](https://tex.z-dn.net/?f=%5Cfrac%7BR%7D%7B4.0%5Ctimes%2010%5E4%7D%3D%5Cfrac%7Bk%28%5Cfrac%7B%5BH_2%5D%7D%7B2%7D%29%5E2%5BI_2%5D%5E2%7D%7Bk%5BH_2%5D%5E2%5BI_2%5D%5E2%7D)
Therefore, the initial rate of the reaction will be,
Answer:
Bones are the correct answer.
Control is something that remains the same or constant through an experiment. And an experiment group is the group you test the one change on. (Also called the dependent variable)
Answer:
ΔG°rxn = -69.0 kJ
Explanation:
Let's consider the following thermochemical equation.
N₂O(g) + NO₂(g) → 3 NO(g) ΔG°rxn = -23.0 kJ
Since ΔG°rxn < 0, this reaction is exergonic, that is, 23.0 kJ of energy are released. The Gibbs free energy is an extensive property, meaning that it depends on the amount of matter. Then, if we multiply the amount of matter by 3 (by multiplying the stoichiometric coefficients by 3), the ΔG°rxn will also be tripled.
3 N₂O(g) + 3 NO₂(g) → 9 NO(g) ΔG°rxn = -69.0 kJ
