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CaHeK987 [17]
2 years ago
7

You have 125.0mL of a solution of H3PO4, but you don't know its concentration. If you titrate the solution with a 4.56M solution

of NaOH and reach the endpoint when 134.1mL of the base are added, what is the concentration of the acid?

Chemistry
1 answer:
Mariulka [41]2 years ago
5 0
4.89 mol/L

You can check the attached image for work.

-T.B.

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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
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
If you reacted 183 grams of copper sulfate with excess iron, what mass of copper would you expect to make? You may need to balan
Nezavi [6.7K]

The question does not provide the equation

Answer:-

72.89 grams

Explanation:-

The balanced chemical equation for this reaction is

CuSO4 + Fe --> FeSO4 + Cu

Molecular weight of CuSO4 = 63.55 x 1 + 32 x 1 + 16 x 4

= 159.55 gram

Atomic weight of Cu = 63.55 gram.

According to the balanced chemical equation

1 CuSO4 gives 1 Cu

∴159.55 gram of CuSO4 would give 63.55 gram of Cu.

183 gram of CuSO4 would give 63.55 x 183 / 159.55

= 72.89 grams of Cu

7 0
3 years ago
In general, what causes a sea breeze?
ella [17]

Answer:

The usual cause of sea breeze is the difference in specific heat capacity between land and water Land heats and cools more quickly than water. What is the cause of a sea breeze? The air over the land is warmed and rises, to a greater temperature than that over the sea.

Explanation:

Your answer is C ! <3

7 0
3 years ago
Ldentify the atom whose nucleus is<br> composed of 56 protons and 82<br> neutrons.
kenny6666 [7]

Answer:

Barium-138

Explanation:

The mass number is the atomic number + the number of neutrons. ... Identify the atom whose nucleus is composed of 56 protons and 82 neutrons. Barium-138 (the number of protons identifies the atom) Gallium has two naturally occurring isotopes: Ga-69 with a mass of 68.9256 amu and a natural abundance of 60.11% and Ga-71.

4 0
2 years ago
1. In this experiment, what property of NaCl is used to separate it from the other two components? Is this a chemical or physica
irinina [24]

Answer 1) In the mixture of sand, NaCl and CaCO_{3} first separation was done by dissolving the mixture in water. In the first step the NaCl will get dissolved whereas, CaCO_{3} will be undissolved and sand will settle at the bottom. Here, the physical property of solubility of NaCl in water is taken into consideration. The collected water is then filtered off and evaporated to get the NaCl back from the mixture with some loss. No chemical change occurs in case of NaCl extraction from water.


Answer 2) When CaCO_{3} was to be removed from the undissolved part. The chemical property of CaCO_{3} was used. Where it gets dissolved in acidic medium. And this way we can extract CaCO_{3} and remove sand from it. We change the chemical composition of CaCO_{3} by adding HCl to the mixture and dissolve CaCO_{3} to form CaCl_{2} which gets dissolved into the solution of HCl. Here, first decantation occurs and then extraction is done. Second, extraction is done using potassium carbonate in it which separates CaCl_{2} from sand.


Answer 3) Here, Unknown mixture of salt and sand weighed =7.52 g (before washing);


Unknown mixture of salt and sand weighed =3.45 g (after washing);


To calculate the amount of salt in it, we can simply subtract the values of before and after washing change in weights.


The property of NaCl being soluble in water it will go away with washing leaving behind the sand only, after washing.


So, Weight of salt was = 7.52g - 3.45 g = 4.07g


To find the percentage of sand that was mixed with salt =

(3.45 g sand / 7.52g of mixture) X 100 = 45.9% sand was mixed with salt.


To verify whether the correct percentage of salt and sand was calculated we can recheck the value for salt as well.

(4.07 g of salt / 7.52g of mixture) X 100 = 54.1% salt


On adding we get, 45.9% + 54.1% = 100%.


Which confirms that the calculations are correct.

4 0
3 years ago
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