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vovikov84 [41]
3 years ago
12

What might you expect to observe from this double-displacement reaction: CoCl2(aq) + Na2SO4(aq) ---> CoSO4 + 2NaCl?

Chemistry
1 answer:
umka21 [38]3 years ago
7 0
To decide the outcome of this reaction, we need to know the state of the products, as in if they are solid, liquid, gaseous, aqueous, etc. The proper reaction is as follows:

CoCl₂ (aq) + Na₂SO₄ (aq) → CoSO₄ (aq) + 2NaCl (aq)

If we were to look up the species CoSO₄ and NaCl, we would find that both of these species are also aqueous. Therefore, neither CoSO₄ or NaCl will precipitate from the solution, so we can ignore options A and B. Neither species formed is gaseous either, therefore, it cannot be option C. That leaves us with option D as the answer.

D is the correct answer because the product of the double displacement is the formation of two more aqueous species which will remain dissociated in the aqueous solution leading to no chemical reaction overall.
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Would the following configuration represent
makvit [3.9K]

Answer:

Excited state

Explanation:

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3 years ago
One isotope of Br has a half-life of 16.5 hours. How much of a 2.00 gram sample remains at the end of 1.00 day?
valina [46]

Answer:

half life=16.5hrs

sample=2g

total time=1day=24hrs

no. of half lives=total time/half life

no. of half lives=24hrs/16.5hrs=1.45approx1.5

sample left=1/2n=1/2[1.5]=0.707gapprox

Explanation:

4 0
3 years ago
A geochemist in the field takes a 36.0 mL sample of water from a rock pool lined with crystals of a certain mineral compound X.
NeTakaya

Answer:

solubility of X in water at 17.0 ^{0}\textrm{C} is 0.11 g/mL.

Explanation:

Yes, the solubility of X in water at 17.0 ^{0}\textrm{C} can be calculated using the information given.

Let's assume solubility of X in water at 17.0 ^{0}\textrm{C} is y g/mL

The geochemist ultimately got 3.96 g of crystals of X after evaporating the diluted solution made by diluting the 36.0 mL of stock solution.

So, solubility of X in 1 mL of water = y g

Hence, solubility of X in 36.0 mL of water = 36y g

So, 36y = 3.96

   or, y = \frac{3.96}{36} = 0.11

Hence solubility of X in water at 17.0 ^{0}\textrm{C} is 0.11 g/mL.

3 0
3 years ago
What is the volume of a gas if 0.182 moles of the gas is at 1.99 atm and 83.4oC?
Pie

\text{Given that,}\\\\\text{Pressure,}~ P = 1.99~ \text{atm}\\ \\\text{No. of moles,}~ n = 0.182~ \text{mol.}\\\\\text{Temperature,}~ T = 83.4^{\circ}~  C = 356.4~K\\\\ \text{Molar gas constant,}~ R = 0.082~ \text{L}~ \text{atm}~ \text{mol}^{-1}~ \text{K}^{-1}\\\\\text{Volume,}~ V =?\\\\\text{We know that,}~\\\\~~~~~~~PV =nRT\\\\\implies V = \dfrac{nRT}{P}\\\\\\~~~~~~~~~~=\dfrac{0.182 \times0.082 \times 356.4 }{1.99}\\\\\\~~~~~~~~~~\approx 2.68~ \text{L.}\\\\\\

\text{Hence the volume of the  gas is 2.68 L.}

6 0
1 year ago
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Your answer is false
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3 years ago
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