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adoni [48]
3 years ago
15

which of these pieces of equipment would be the most appropriate for precisely measuring 29 mL of liquid? Explain your reasoning

, citing evidence related to the divisions on each tool and the concept of an estimated digit.​

Chemistry
1 answer:
Anna007 [38]3 years ago
5 0

Answer:

The best equipment would be the graduated cylinder. Why?

Firstly, the smallest marking on the graduated cylinder is 2 mL, while on all the others the smallest marking is way above that, like 25 mL and 100 mL.

Without even going into the details, we can first rule out the volumetric flask, since its smallest marking is 100 mL and even that is already bigger than our sample size, hence we would have no markings to accurately measure out 29 mL of our sample had we used the volumetric flask.

Next to be ruled out would be the Erlenmeyer flask, as you can see in the image, it only has three marking, and as the smallest marking is 25 mL, each marking is at least 25 mL, and even so far as going up to 50 mL. This cannot let us accurately measure 29 mL out at all, due to the markings being way too big to do that. Hence, the Erlenmeyer flask is ruled out.

Finally, the beaker seems to be a worthy candidate! Unfortunately, for the same reason as the Erlenmeyer flask, as you can see in the image each marking represents 10 mL. We cannot measure 9 mL in the beaker accurately, and hence the beaker is ruled too.

We are left with the graduated cylinder, and that is our answer.

Explanation:

Hope this helped!

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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:

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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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We know, density=\frac{mass}{volume}

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