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sveticcg [70]
3 years ago
9

Calculate the percent composition by mass of iron in Fe(NO3)3

Chemistry
1 answer:
Dovator [93]3 years ago
6 0

Answer:

It's 23.14 percent

Explanation:

First, the mass of all the elements are:

N = 14

O = 16

Fe = 56

In this molecule you have 3 atoms of N, and 9 atoms of O, so:

3•14 = 42

16•9 = 144

The whole mass of the molecule is:

56 + 42 + 144 = 242

242/100 = 2.42, so 1% is 2.42

56/2.42 = 23.14%

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Explanation:

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Omar wrote a hypothesis about batteries called dry cells.
Flura [38]
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3. If the percent by volume is 2.0% and the volume of solution is 250 mL, what is the volume of solute in solution? (1 point) 0.
igor_vitrenko [27]
Volume percent = Volume of solute
                              ----------------------------------
                                Volume of the solution
  
                  2                    Volume of the solute
               -------   =           ------------------------------
                100                               250
                     
         Volume of the solute = 2 x 250
                                                ------------
                                                  100         

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4 0
3 years ago
What super power would you want to have and why?<br><br> I will give Brainly to the best answer
Darya [45]

Answer:

If I could have a super power it would be invisibility. Sometimes you wish you weren't there and if you were hiding from someone then they couldn't find you.

3 0
2 years ago
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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