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Stella [2.4K]
2 years ago
15

When K2SO4 is separated into its ions, how is it written?

Chemistry
2 answers:
VashaNatasha [74]2 years ago
8 0
When K2SO4 is separated into its ions, it is written like this.

K2SO4 -> 2K+ + SO4^-2.

This is a balanced equation.
Neko [114]2 years ago
8 0

Answer: K^+ and SO_4^-^2

Explanation: Potassium is a first group metal with one valence electron and so it has +1 charge. Sulfate ion is a polyatomic ion with -2 charge. The compound potassium sulfate has two potassium ions and one sulfate ion and the below equation shows how the ions are written when the compound is dissociated into its ions.

K_2SO_4\rightarrow 2K^++SO_4^-^2

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The ionic equations represent the reactions between four metals, P, Q, R and S, and solutions of the salts of the same metals
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Answer:

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5 0
3 years ago
A chemist wishing to do an experiment requiring 47-Ca2+ (half-life = 4.5 days) needs 5.0 μg of the nuclide. What mass of 47-CaCO
NARA [144]

Answer:

5.8μg

Explanation:

According to the rate or decay law:

N/N₀ = exp(-λt)------------------------------- (1)

Where N = Current quantity,  μg

            N₀ = Original quantity, μg

             λ= Decay constant day⁻¹

              t =  time in days

Since the half life is 4.5 days, we can calculate the  λ from (1) by  substituting N/N₀ = 0.5

0.5 = exp (-4.5λ)

ln 0.5  = -4.5λ

-0.6931 = -4.5λ

λ =   -0.6931 /-4.5

  =0.1540 day⁻¹

Substituting into (1)  we have :

N/N₀ = exp(-0.154t)----------------------------- (2)

To receive 5.0 μg of the nuclide with a delivery time of 24 hours or 1 day:

N = 5.0 μg

N₀ = Unknown

t = 1 day

Substituting into (2) we have

[5/N₀]   = exp (-0.154 x 1)

    5/N₀        = 0.8572

N₀  =  5/0.8572

     =    5.8329μg

    ≈     5.8μg

The Chemist must order 5.8μg  of 47-CaCO3

6 0
2 years ago
Explain how chemists overcome the problem of low yield In industry
mrs_skeptik [129]

High temperature and pressure produce the highest rate of reaction. However, this must be balanced with the high cost of the energy needed to maintain these conditions. Catalysts increase the rate of reaction without affecting the yield. This can help create processes which work well even at lower temperatures.

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