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bija089 [108]
3 years ago
7

1.88gof potassium chloride is dissolved in300.mLof a23.0mMaqueous solution of silver nitrate.Calculate the final molarity of chl

oride anion in the solution. You can assume the volume of the solution doesn't change when the potassium chloride is dissolved in it.Be sure your answer has the correct number of significant digits.
Chemistry
1 answer:
Dvinal [7]3 years ago
5 0

Answer:

The final molarity of chloride anion in the solution is 0 molar.

Explanation:

Mass of potassium chloride solution = 1.88 g

Moles of potassium chloride = \frac{1.88 g}{74.5 g/mol}=0.02524 mol

Moles of silver nitrate = n

Molarity of silver nitrate solution = M= 23.0 M

Volume of the silver nitrate solution ,V= 300.0 mL = 0.3 L

moles(n)=Molarity(M)\times Volume (L)

n=23.0 M\times 0.3 L = 6.9 mol

AgNO_3(aq)+KCl(aq)\rightarrow AgCl(s)+KNO_3(aq)

According to reaction 1 mol of potassium nitrate reacts with 1 mol of silver nitrate .

Then 0.02524 moles of potassium chloride will react with:

\frac{1}{1}\times 0.02524 mol=0.02524 mol of silver nitrate.

As we can see that potassium chloride is in limiting amount due to which the potassium chloride will get completely converted into silver chloride and potassium nitrate.

Since , no potassium chloride will left after reaction which indicates that no chloride ions will be present after reaction.

[Cl^-]=0 M

The final molarity of chloride anion in the solution is 0 molar.

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First we <u>convert 1.2 x 10²³ atoms of arsenic (As) into moles</u>, using <em>Avogadro's number</em>:

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Then we can<u> calculate the mass of 0.199 moles of arsenic</u>, using its<em> molar mass</em>:

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6 0
3 years ago
If this decay has a half life of 2.60 years, what mass of 72.5 g of sodium 22 will remain after 15.6 years
Vlad1618 [11]

Sodium-22 remain : 1.13 g

<h3>Further explanation </h3>

The atomic nucleus can experience decay into 2 particles or more due to the instability of its atomic nucleus.  

Usually, radioactive elements have an unstable atomic nucleus.  

General formulas used in decay:  

\large{\boxed{\bold{N_t=N_0(\dfrac{1}{2})^{T/t\frac{1}{2} }}}

T = duration of decay  

t 1/2 = half-life  

N₀ = the number of initial radioactive atoms  

Nt = the number of radioactive atoms left after decaying during T time  

half-life = t 1/2=2.6 years

T=15.6 years

No=72.5 g

\tt Nt=72.5.\dfrac{1}{2}^{15.6/2.6}\\\\Nt=72.5.\dfrac{1}{2}^6\\\\Nt=1.13~g

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3 years ago
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