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Advocard [28]
1 year ago
12

The reaction NA3PO4(aq) + 3 AgNO3(aq) → Ag3PO4() + 3NaNO 3( aq) is best classified as a(n)?

Chemistry
1 answer:
Natasha2012 [34]1 year ago
8 0

Answer:

Double replacement reaction.

Explanation:

The Na and Ag atoms both (double) trade places (replacement) with each other.  

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Color change
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a technical machinist is asked to build a cubical steel tank that will hold 270L of water. Calculate in meters the smallest poss
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length of tank = 0.65m

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In a real system of levers, wheels, or pulleys, the AMA is less than the IMA because _____.
kodGreya [7K]

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the work input is depented on the work output

Explanation:

7 0
3 years ago
An ion with a -2 charge has? A) one missing protein B) two missing protons c) two missing electrons D) two extra electrons
Andreyy89

Answer:

D.

Explanation:

It would not be gain because it would say +6 instead of -2. Having a -2 charge means it wants to loose 2 electrons so it can form a full shell of 8 electrons. It wants to be inert (stable).

- Hope that helps! Please let me know if you need further explanation.

3 0
2 years ago
Read 2 more answers
From the value Kf=1.2×109 for Ni(NH3)62+, calculate the concentration of NH3 required to just dissolve 0.016 mol of NiC2O4 (Ksp
Nina [5.8K]

<u>Answer:</u> The concentration of NH_3 required will be 0.285 M.

<u>Explanation:</u>

To calculate the molarity of NiC_2O_4, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Moles of NiC_2O_4 = 0.016 moles

Volume of solution = 1 L

Putting values in above equation, we get:

\text{Molarity of }NiC_2O_4=\frac{0.016mol}{1L}=0.016M

For the given chemical equations:

NiC_2O_4(s)\rightleftharpoons Ni^{2+}(aq.)+C_2O_4^{2-}(aq.);K_{sp}=4.0\times 10^{-10}

Ni^{2+}(aq.)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K_f=1.2\times 10^9

Net equation: NiC_2O_4(s)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K=?

To calculate the equilibrium constant, K for above equation, we get:

K=K_{sp}\times K_f\\K=(4.0\times 10^{-10})\times (1.2\times 10^9)=0.48

The expression for equilibrium constant of above equation is:

K=\frac{[C_2O_4^{2-}][[Ni(NH_3)_6]^{2+}]}{[NiC_2O_4][NH_3]^6}

As, NiC_2O_4 is a solid, so its activity is taken as 1 and so for C_2O_4^{2-}

We are given:

[[Ni(NH_3)_6]^{2+}]=0.016M

Putting values in above equations, we get:

0.48=\frac{0.016}{[NH_3]^6}}

[NH_3]=0.285M

Hence, the concentration of NH_3 required will be 0.285 M.

7 0
3 years ago
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