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Reil [10]
3 years ago
7

Please help me ill give brainiest

Chemistry
2 answers:
V125BC [204]3 years ago
6 0
Reaction on the left product on the right
lyudmila [28]3 years ago
5 0
Your reactants are on the left side and your product is on the right side
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In the equation shown, what are the reactant(s)?
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C3H8 + O2 (please give me brainliest$
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2 years ago
Predict the sign of the entropy change,Delta S, for each of the following reactions:The signs are either going to be pos or nega
algol [13]

Answer: a) Pb^{2+}(aq)+2Cl^-(aq)\rightarrow PbCl_2(s):  negative

b)  CaCO_3(s)\rightarrow CaO(s)+CO_2(g) : positive

c) 2NH_3(g)\rightarrow N_2(g)+3H_2(g): positive.

d) P_4(g)+5O_2(g)\rightarrow P_4O_{10}(s) : negative

e) C_4H_8(g)+6O_2(g)\rightarrow 4CO_2(g)+4H_2O(g):  positive.

f) I_2(s)\rightarrow I_2(g) : positive.

Explanation:

Entropy is the measure of randomness or disorder of a system. If a system moves from  an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.

\Delta S is positive when randomness increases and \Delta S is negative when randomness decreases.

a) Pb^{2+}(aq)+2Cl^-(aq)\rightarrow PbCl_2(s)

As ions are moving to solid form , randomness decreases and thus sign of \Delta S is negative.

b) CaCO_3(s)\rightarrow CaO(s)+CO_2(g)

As solid is changing to gas, randomness increases and thus sign of \Delta S is positive.

c) 2NH_3(g)\rightarrow N_2(g)+3H_2(g)

As 2 moles of reactants are converted to 4 moles of products , randomness increases and thus sign of \Delta S is positive.

d) P_4(g)+5O_2(g)\rightarrow P_4O_{10}(s)

As gas is changing to solid, randomness decreases and thus sign of \Delta S is negative.

e) C_4H_8(g)+6O_2(g)\rightarrow 4CO_2(g)+4H_2O(g)

As 7 moles of reactants are converted to 8 moles of products , randomness increases and thus sign of \Delta S is positive.

f) I_2(s)\rightarrow I_2(g)

As solid is changing to gas, randomness increases and thus sign of \Delta S is positive.

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