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zhuklara [117]
3 years ago
13

50 POINTS*** Which of the following is not a correct chemical equation for a double displacement reaction?

Chemistry
2 answers:
Ira Lisetskai [31]3 years ago
7 0

Answer:-

B) CaCl2 + Li2CO3 --> CaCO3 + 2LiCl

D) 2MgI2 + Mn(SO3)2 ---> 2MgSO3 + MnI4

Explanation:-

For a reaction to be double displacement reaction there are two things we need to look for

1) There must be an interchange of the group of ions

2) The reactants must dissolve in water to release ions

In this question every option shows an interchange of ions. So we must use the second criteria to find out which one is not a double displacement reaction.

In case of B the two reactants are CaCl2 and Li2CO3.

Li2CO3 does not form ions on dissolving in water. It is because carbonate salts are not soluble in water.

Hence B) CaCl2 + Li2CO3 --> CaCO3 + 2LiCl is not a double displacement reaction.

In case of D the two reactants are MgI2 and Mn(SO3)2.

Mn(SO3)2 does not form ions on dissolving in water. It is because sulphite salt of metals except potassium and sodium are not soluble in water.

Hence D) 2MgI2 + Mn(SO3)2 ---> 2MgSO3 + MnI4 is not a double displacement reaction.

hammer [34]3 years ago
4 0
I mostly believe in between D and B beacuse K3po4 and caco3 is not an element equation

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5 0
3 years ago
Si una masa dada de hidrógeno ocupa 40 litros a 700 grados torr. ¿Qué volumen ocupará a 1 atmósfera de presión? (dar la presión
Svetach [21]

Answer:

V_2=36.84L

Explanation:

Hola,

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P_1V_1=P_2V_2

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Saludos!

7 0
3 years ago
Read 2 more answers
Calculate the entropy change for the surroundings of the reaction below at 350K: N2(g) + 3H2(g) -> 2NH3(g) Entropy data: NH3
krek1111 [17]

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).

\Delta S^o=S_{product}-S_{reactant}

\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)}]

where,

\Delta S^o = entropy of reaction = ?

n = number of moles

\Delta S^0{(NH_3)} = standard entropy of NH_3

\Delta S^0{(H_2)} = standard entropy of H_2

\Delta S^0{(N_2)} = standard entropy of N_2

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)]

\Delta S^o=-198.3J/K

Therefore, the entropy change for the surroundings of the reaction is, -198.3 J/K

4 0
2 years ago
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