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Bas_tet [7]
3 years ago
9

What determines the identity of the salt that forms in a neutralization reaction?

Chemistry
1 answer:
nekit [7.7K]3 years ago
6 0
The identities of the acid and base.
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Consider the reaction.
N76 [4]

Answer: m = 50 g ZnSO4

Explanation: First is convert the moles of Zn to the moles of ZnSO4 by having their mole ratio which is 2:2 based from the balanced equation. Next is convert the moles of ZnSO4 to mass using its molar mass.

0.311 mole Zn x 2 moles ZnSO4 / 2 moles Zn

= 0.311 moles ZnSO4

0.311 moles ZnSO4 x 161 g ZnSO4 / 1 mole ZnSO4

= 50 ZnSO4

5 0
3 years ago
Read 2 more answers
The unit cell of amazonite is made of three pairs of faces, each pair different from the others. The faces are oriented at angle
scZoUnD [109]
<span>The unit cell of amazonite is made of three pairs of faces, each pair different from the others. The faces are oriented at angles to one another that are not 90 degrees. The type of crystal this is is triclinic.
Triclinic crystal system is one of the seven crystal systems, with three basis vectors. </span>
5 0
2 years ago
Calculate the maximum amount of useful work that can be obtained and comment on the spontaneity for the reaction at 25C :
harkovskaia [24]

Answer:

1.41 *10^{3}  kJ/mol

Explanation:

First, we find in the tables the ΔH of formation of each compound. As you can see in the (image 1)

Then we solve the ecuation for ΔH°reaction

ΔH°reaction=∑ΔH°f(products)−∑ΔH°f(Reactants)

ΔH°reaction= (-2* 393.5 - 2*285.8) - (52.4 + 0) kJ/mol

ΔH°reaction = -1.41 *10^3  kJ/mol

3 0
3 years ago
Which of the following is a true statement about the water cycle?
Serggg [28]
I think its C: due to mixing and oceanic.....
3 0
3 years ago
A nuclear waste site. cesium-137 is a particularly dangerous by-product of nuclear reactors. it has a half-life of 30 years. it
hram777 [196]
The mass decay rate is of the form
m(t) = m_{0} e^{-kt}
where
m₀ = 3000 g,the initial mass
k = the decay constant
t = time, years.

Because the half-life is 30 years, therefore
e^{-30k} =  \frac{1}{2} \\\ -30k = ln(0.5) \\ k =  \frac{ln(0.5)}{-30} =0.0231

After 60 years, the mass remaining is
m = 3000 e^{-0.0231*60} = 750 \, g

Answer: 750 g

4 0
3 years ago
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