3 Common States of Matter:
1. Solid - particles are motionless and stick together very closely.
2. Liquid - particles are moving slowly without pattern.
3. Gas - Particles are moving rapidly again without pattern.
Answer: Hợp chất CTHH 0 °C 10 °C 20 °C 30 °C 40 °C 50 °C 70 °C
Actini(III) hydroxide Ac(OH)3 0,0022
Amonia NH3 1176 900 702 565 428 333 188
Amoni azua NH4N3 16 25,3 37,1
View 42 more rows
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<u>Answer:</u>
Exothermic Reaction are those reaction, in which energy is released while in endothermic reaction are those, in which energy is absorbed.
<u>Explanation:</u>
First Reaction:
As in this reaction, energy is released
½H2(g) + ½I2(g) → HI(g), ΔH = +6.2 kcal/mole
so it is <em>exothermic reaction</em>
Second reaction:
As in this reaction, energy is absorbed
21.0 kcal/mole + C(s) + 2S(s) → CS2(l)
so it is <em>endothermic reactions</em>.
A) solid
b)liquid
c)liquid
d)gas
Answer : The entropy change for the surroundings of the reaction is, -198.3 J/K
Explanation :
We have to calculate the entropy change of reaction 
.
![\Delta S^o=[n_{NH_3}\times \Delta S^0_{(NH_3)}]-[n_{N_2}\times \Delta S^0_{(N_2)}+n_{H_2}\times \Delta S^0_{(H_2)}]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D%5Bn_%7BNH_3%7D%5Ctimes%20%5CDelta%20S%5E0_%7B%28NH_3%29%7D%5D-%5Bn_%7BN_2%7D%5Ctimes%20%5CDelta%20S%5E0_%7B%28N_2%29%7D%2Bn_%7BH_2%7D%5Ctimes%20%5CDelta%20S%5E0_%7B%28H_2%29%7D%5D)
where,
= entropy of reaction = ?
n = number of moles
= standard entropy of 
= standard entropy of 
= standard entropy of 
Now put all the given values in this expression, we get:
![\Delta S^o=[2mole\times (192.5J/K.mole)]-[1mole\times (191.5J/K.mole)+3mole\times (130.6J/K.mole)]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D%5B2mole%5Ctimes%20%28192.5J%2FK.mole%29%5D-%5B1mole%5Ctimes%20%28191.5J%2FK.mole%29%2B3mole%5Ctimes%20%28130.6J%2FK.mole%29%5D)
Therefore, the entropy change for the surroundings of the reaction is, -198.3 J/K
