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const2013 [10]
3 years ago
9

What is the simplest formula of a compound if a sample of the compound contains 0.221 mol X, 0.442 mol Y, and 0.884 mol Z? chemP

adHelp XY2Z4 XY_2Z_4 Correct. How many moles of Z would be in a sample that contained 0.588 mol X? WebAssign will check your answer for the correct number of significant figures.
Chemistry
2 answers:
bezimeni [28]3 years ago
7 0

Answer:

The simplest formula of this compound is XY2Z4

We would have 2.352 moles of Z in a sample that contains 0.588 moles X

Explanation:

Step 1: Data given

A compound contains 3 elements

X = 0.221 mol

Y = 0.442 mol

Z = 0.884 mol

Step 2: Calculate the mol ratio

We divide by the smallest amount of moles

X = 0.221/0.221 = 1

Y = 0.442/0.221 = 2

Z = 0.884/0.221 = 4

The empirical formula is XY2Z4

This means for each 1 X atom we have 2 Y atoms and 4 Z atoms.

If we have 0.588 moles X this means we have 0.588*2 = 1.176 moles of Y

And 0.588 *4 = 1.176*2 = 2.352 moles of Z

The simplest formula of this compound is XY2Z4

We would have 2.352 moles of Z in a sample that contains 0.588 moles X

Gnoma [55]3 years ago
6 0

Answer:

XY₂Z₄

2.35 mol Z

Explanation:

A sample of the compound contains 0.221 mol X, 0.442 mol Y, and 0.884 mol Z. We can find the simplest formula (empirical formula) by <em>dividing all the numbers of moles by the smallest one</em>.

X: 0.221/0.221 = 1

Y: 0.442/0.221 = 2

Z: 0.884/0.221 = 4

The simplest formula is XY₂Z₄.

The molar ratio of X to Z is 1:4. The moles of Z in a sample that contained 0.588 moles of X is:

0.588 mol X × (4 mol Z/1 mol X) = 2.35 mol Z

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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
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<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.

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