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tatuchka [14]
2 years ago
5

What is the balanced NET ionic equation for the reaction when aqueous Cs₃PO₄ and aqueous AgNO₃ are mixed in solution to form sol

id Ag₃PO₄ and aqueous CsNO₃
Chemistry
1 answer:
rusak2 [61]2 years ago
7 0

Answer:

PO_4^{3-}(aq)+3Ag^+(aq)\rightarrow Ag_3PO_4(s)

Explanation:

Hello!

In this case, since the net ionic equation of a chemical reaction shows up the ionic species that result from the simplification of the spectator ions, which are those at both reactants and products sides, we take into account that aqueous species ionize into ions whereas liquid, solid and gas species remain unionized. In such a way, for the reaction of cesium phosphate and silver nitrate we can write the complete molecular equation:

Cs_3PO_4(aq)+3AgNO_3(aq)\rightarrow Ag_3PO_4(s)+3CsNO_3(aq)

Whereas the three aqueous salts are ionized in order to write the following complete ionic equation:

3Cs^+(aq)+PO_4^{3-}(aq)+3Ag^+(aq)+3NO_3^-(aq)\rightarrow Ag_3PO_4(s)+3Cs^+(aq)+3NO_3^-(aq)

In such a way, since the cesium and nitrate ions are the spectator ions because of the aforementioned, the net ionic equation turns out:

PO_4^{3-}(aq)+3Ag^+(aq)\rightarrow Ag_3PO_4(s)

Best regards!

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Answer:

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oksano4ka [1.4K]

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a) homogeneous : the boundary of the two components is not physically distinct

b) heterogeneous:the boundary of the two components is  physically distinct

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1) filtration: if the two components are forming heterogeneous mixture we can separate them by filtration.

2) boiling: if boiling point of one of the components is less than other

3) magnetic separation: if one of the component is magnetic

4)sieve method: for solid components with difference in size of particles

5) hand picking

Thus the correct match will be as shown in the figure


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A balloon is filled with 0.250 mole of air at 35°C. If the volume of the balloon is 6.23 liters, what is the absolute pressure o
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Which is an example of a double
Ostrovityanka [42]

Answer:

NaCl + AgF → NaF + AgCl

Explanation:

A double replacement reaction is a type of chemical reaction that occurs when two reactants exchange cations or anions to yield two new products.

From all the reactions given ,

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6 0
3 years ago
Since the half-life of 235U (7. 13 x 108 years) is less than that of 238U (4.51 x 109 years), the isotopic abundance of 235U has
Ymorist [56]

Answer:

\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}

Explanation:

Given that:

The Half-life of ^{235}U = 7.13 \times 10^8 \ years is less than that of ^{238} U = 4.51 \times 10^9 \ years

Although we are not given any value about the present weight of ^{235}U.

So, consider the present weight in the percentage of ^{235}U to be  y%

Then, the time elapsed to get the present weight of ^{235}U = t_1

Therefore;

N_1 = N_o e^{-\lambda \ t_1}

here;

N_1 = Number of radioactive atoms relating to the weight of y of ^{235}U

Thus:

In( \dfrac{N_1}{N_o}) = - \lambda t_1

In( \dfrac{N_o}{N_1}) =  \lambda t_1 --- (1)

However, Suppose the time elapsed from the initial stage to arrive at the weight of the percentage of ^{235}U to be = t_2

Then:

In( \dfrac{N_o}{N_2}) =  \lambda t_2  ---- (2)

here;

N_2 =  Number of radioactive atoms of ^{235}U relating to 3.0 a/o weight

Now, equating  equation (1) and (2) together, we have:

In( \dfrac{N_o}{N_1}) -In( \dfrac{N_o}{N_2}) =  \lambda( t_1-t_2)

replacing the half-life of ^{235}U = 7.13 \times 10^8 \ years

In( \dfrac{N_2}{N_1})  = \dfrac{In 2}{7.13 \times 10^9}( t_1-t_2)      ( since \lambda = \dfrac{In 2}{t_{1/2}} )

∴

\mathtt{In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2}= t_1-t_2}

The time elapsed signifies how long the isotopic abundance of 235U equal to 3.0 a/o

Thus, The time elapsed is  \mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}

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