Answer:
When iron sulfide is exposed to oxygen it reacts rapidly, releasing large amounts of heat.
Explanation:
i took this from my notes- so whatever aligns with that. sorry if this didnt help but i think its genuinely c heating
Answer:
Rate of reaction =
Rate of consumption of A = 
Rate of consumption of B =
Rate of formation of D = 
Explanation:
According to laws of mass action for the given reaction,
![Rate= -\frac{1}{2}\frac{\Delta [A]}{\Delta t}=-\frac{\Delta [B]}{\Delta t}=\frac{1}{2}\frac{\Delta [C]}{\Delta t}=\frac{1}{3}\frac{\Delta [D]}{\Delta t}](https://tex.z-dn.net/?f=Rate%3D%20-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7B%5CDelta%20%5BA%5D%7D%7B%5CDelta%20t%7D%3D-%5Cfrac%7B%5CDelta%20%5BB%5D%7D%7B%5CDelta%20t%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7B%5CDelta%20%5BC%5D%7D%7B%5CDelta%20t%7D%3D%5Cfrac%7B1%7D%7B3%7D%5Cfrac%7B%5CDelta%20%5BD%5D%7D%7B%5CDelta%20t%7D)
where, ![-\frac{\Delta [A]}{\Delta t}](https://tex.z-dn.net/?f=-%5Cfrac%7B%5CDelta%20%5BA%5D%7D%7B%5CDelta%20t%7D)
is rate of consumption of A, ![-\frac{\Delta [B]}{\Delta t}](https://tex.z-dn.net/?f=-%5Cfrac%7B%5CDelta%20%5BB%5D%7D%7B%5CDelta%20t%7D)
is rate of consumption of B, ![\frac{\Delta [C]}{\Delta t}](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%20%5BC%5D%7D%7B%5CDelta%20t%7D)
is rate of formation of C and ![\frac{\Delta [D]}{\Delta t}](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%20%5BD%5D%7D%7B%5CDelta%20t%7D)
is rate of formation of D
Here ![\frac{\Delta [C]}{\Delta t}=2.7mol.dm^{-3}.s^{-1}](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%20%5BC%5D%7D%7B%5CDelta%20t%7D%3D2.7mol.dm%5E%7B-3%7D.s%5E%7B-1%7D)
So, Rate of reaction = 
Rate of formation of D = ![(\frac{3}{2}\times \frac{\Delta [C]}{\Delta t})=(\frac{3}{2}\times 2.7mol.dm^{-3}.s^{-1})=4.15mol.dm^{-3}.s^{-1}](https://tex.z-dn.net/?f=%28%5Cfrac%7B3%7D%7B2%7D%5Ctimes%20%5Cfrac%7B%5CDelta%20%5BC%5D%7D%7B%5CDelta%20t%7D%29%3D%28%5Cfrac%7B3%7D%7B2%7D%5Ctimes%202.7mol.dm%5E%7B-3%7D.s%5E%7B-1%7D%29%3D4.15mol.dm%5E%7B-3%7D.s%5E%7B-1%7D)
Rate of consumption of A = ![(\frac{2}{2}\times \frac{\Delta [C]}{\Delta t})=(\frac{2}{2}\times 2.7mol.dm^{-3}.s^{-1})=2.7mol.dm^{-3}.s^{-1}](https://tex.z-dn.net/?f=%28%5Cfrac%7B2%7D%7B2%7D%5Ctimes%20%5Cfrac%7B%5CDelta%20%5BC%5D%7D%7B%5CDelta%20t%7D%29%3D%28%5Cfrac%7B2%7D%7B2%7D%5Ctimes%202.7mol.dm%5E%7B-3%7D.s%5E%7B-1%7D%29%3D2.7mol.dm%5E%7B-3%7D.s%5E%7B-1%7D)
Rate of consumption of B = ![(\frac{1}{2}\times \frac{\Delta [C]}{\Delta t})=(\frac{1}{2}\times 2.7mol.dm^{-3}.s^{-1})=1.35mol.dm^{-3}.s^{-1}](https://tex.z-dn.net/?f=%28%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%5Cfrac%7B%5CDelta%20%5BC%5D%7D%7B%5CDelta%20t%7D%29%3D%28%5Cfrac%7B1%7D%7B2%7D%5Ctimes%202.7mol.dm%5E%7B-3%7D.s%5E%7B-1%7D%29%3D1.35mol.dm%5E%7B-3%7D.s%5E%7B-1%7D)
AgNO3 + Cu = Cu(NO3)2 + 2Ag .. this is an example of a single replacement reaction.
percentage of zinc= 5.985/20 X 100= 29.5%
let x represents zinc and y be magnesium so
Zinc(s) + 2HCl(aq) -----> ZnCl2(aq) + H2(g)
Mg(s) + 2HCl(aq) -----> MgCl2(aq) + H2(g)
1.152 g of hydrogen gas=( x/65.3 + y/24.305) 20
= 3.08 x10^-2 + 0.8295-8.30
=x=5.985
y= 4.015
percentage of zinc= 5.985/20 X 100= 29.5%
The hydrogen gas escapes, so the two products of Zinc Chloride and Magnesium Chloride will then react with Silver Nitrate. Again, I combine both of the reactions
To learn more about zinc:
brainly.com/question/13890062
#SPJ4
Answer:
A reaction in which the entropy of the system decreases can be spontaneous only if it is exothermic.
Explanation:
The spontaneity of a reaction depends on the Gibbs free energy(ΔG).
- If ΔG < 0, the reaction is spontaneous.
- If ΔG > 0, the reaction is nonspontaneous.
ΔG is related to the enthalpy (ΔH) and the entropy (ΔS) through the following expression:
ΔG = ΔH - T.ΔS
where,
T is the absolute temperature (always positive)
Regarding the exchange of heat:
- If ΔH < 0, the reaction is exothermic.
- If ΔH > 0, the reaction is endothermic.
<em>Which statement is true? </em>
<em>A reaction in which the entropy of the system decreases can be spontaneous only if it is exothermic. </em>TRUE. If ΔS < 0, the term -T.ΔS > 0. ΔG can be negative only if ΔH is negative.
<em>A reaction in which the entropy of the system increases can be spontaneous only if it is endothermic.</em> FALSE. If ΔS > 0, the term -T.ΔS < 0. ΔG can be negative if ΔH is negative.
<em>A reaction in which the entropy of the system decreases can be spontaneous only if it is endothermic.</em> FALSE. If ΔS < 0, the term -T.ΔS > 0. ΔG cannot be negative if ΔH is positive.
<em>A reaction in which the entropy of the system increases can be spontaneous only if it is exothermic.</em> FALSE. If ΔS > 0, the term -T.ΔS < 0. ΔG can be negative even if ΔH is positive, as long as |T.ΔS| > |ΔH|.
