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Vladimir [108]
3 years ago
10

15.How can a pure sample of barium sulphate be otained from barium carbonate?

Chemistry
1 answer:
yan [13]3 years ago
8 0

Answer:

Barium carbonate powder is stirred add pulp in the entry, the vitriol that the adds solubility then reaction that makes the transition is filtered and is obtained the barium sulfate filter cake and liquid after the transition.

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D=___g/mL<br> v=100mL<br> m=1.5kg=___g
Bess [88]

The volume is 100mL.

The mass is 1.5kg which is equal to 1500g.

Thus, the density is 1500g / 100mL which is 15g/mL.

8 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
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What is refraction? * Bending of light<br> Reflection of light
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Answer:

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Explanation:

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