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3 years ago

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Five drops of 1.0M NaOH increase the pH of a solution from 3.0 to 4.6. Five additional drops of 1.0M NaOH increase the pH to 4.7

. These results are evidence that the solution is buffered with _____.

1 answer:

Andrej [43]3 years ago

5 0

The solution is buffered with a weak acid and it's conjugate base. Since NaOH is a base, you can use a weak acid as it's buffer.

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As an electron gains energy, it becomes excited and jumps to a lower energy level?

Maru [420]

Yes if you add an energy to an electron the electron will become excited, and it will jump to its highest level then go back down releasing energy

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3 years ago

Read 2 more answers

Round off the following number to 3 digits 34,560

Vikki [24]

Answer:

3.46x10⁴

Explanation:

Hello,

In this case, we can see that the number 34,560 has five significant figures, it means that if we want to write it with three, we must take the 3, 4 and 5 only. Nevertheless, since the 6 after the five is greater than 5, we can round such five to 6, so we obtain:

346

However, the decimal places cannot get lost, therefore, we move the given thousand to the three, so the number turns out:

3.46x10⁴

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6 0

3 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.

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3 years ago

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To calculate the mass of Fe formed in a) we get first the limiting reactant between Fe2O3 and CO. Given the masses, the ratio of Fe2O3 is 1.33 while that of CO is 1.67. Hence the limiting reagent is Fe2O3. The mass of Fe formed is 148.98 grams. In b, the needed CO is only 112.04 grams. Hence, the excess is 27. 96 grams.

8 0

3 years ago