When 1 mole of Zn and 2 mole of O₂ reacts togethor, It will produce 1 mole of ZnO as O₂ is excess reagent and Zn will act as Limiting reagent and thus it limits the amount of product formed.
<h3>What is Limiting reagent ?</h3>
Limiting reagents are the substances that are completely consumed first in a chemical reaction.
Given ;
- Amopunt of Zn : 1 mole
- Amount of O₂ : 2 mole
Given equation ;
2Zn + O₂ --> 2ZnO
As according to given chemical equation,
2 moles Zn of reacts with 1 mole of O₂ to produce 2 moles of ZnO.
Therefore,
1 mole of Zn will require 0.5 mole of O₂.
But,
the given amount of O₂ is 2 mole which is excess
Hence,
Zn here will act as the Limiting reagent, According to which the amount of product formed will be decided.
Therefore,
If 2 mole Zn produces : 2 moles of ZnO (According to given balanced equation)
Thus,
1 mole of Zn will produce = 2/2 x 1
= 1 mole of ZnO
Hence, When 1 mole of Zn and 2 mole of O₂ reacts togethor, It will produce 1 mole of ZnO as O₂ is excess reagent and Zn will act as Limiting reagent and thus it limits the amount of product formed.
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The number of years that must be invested at a rate of 7 % to earn$ 303.52 in interest is
8 years
<em><u>calculation</u></em>
- <em><u> </u></em><em>by use of the formula A = P (1+ rt) </em>
- <em> where : A is the final amount = 542 + 303.52 =$ 845.52</em>
<em> P is the principal money to be invested = $ 542</em>
<em> r= rate= 7/100=0.07</em>
<em> t= time required</em>
<em>=$ 845.52=$ 542( 1+ 0.07 t)</em>
- <em>open the bracket</em>
- <em>= $845.53= $542 + $37.94 t</em>
- <em>like terms together</em>
=$ 845.53 -$542 = $37.94 t
=$303.52 =$37.94 t
- divide both side by $37.94
= $303.52/ $ 37.94 = $37.94t/$37.94
t= 8 years
No beginning or end I believe
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|.
