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2 years ago

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A 0.00100 M Na3[Co(CN)] solution with a volume of 50.0 mL was treated with 0.0100 M Hg2(N03)2 to precipitate (Hg2)3[Co(CN)6]2. U

sing activity coefficients, find pCo(CN)6 when 90.0 mL of Hg2(NO3)2 was added.

1 answer:

Fittoniya [83]2 years ago

6 0

Answer:

wreeeeeeeeeeeeeeeeeeeeeeeeeeeee

Explanation:

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If your lungs were filled with air containing this level of lead, how many lead atoms would be in your lungs? (Assume a total lu

kolbaska11 [484]

Answer:

1.505×10^23 atoms of lead

Explanation:

Volume of lead in the lungs = total volume of lungs = 5.60L

1 mole = 22.4L

5.6L of lead = 5.6/22.4 = 0.25 mole

From Avogadro's law

1 mole of lead contains 6.02×10^23 atoms of lead

0.25 mole of lead = 0.25×6.02×10^23 = 1.505×10^23 atoms of lead

6 0

2 years ago

Using the following standard reduction potentials, Fe3+(aq) + e- → Fe2+(aq) E° = +0.77 V Ni2+(aq) + 2 e- → Ni(s) E° = -0.23 V ca

lina2011 [118]

<u>Answer:</u> The above reaction is non-spontaneous.

<u>Explanation:</u>

For the given chemical reaction:

$Ni^{2+}(aq.)+2Fe^{2+}(aq.)\rightarrow 2Fe^{3+}(aq.)+Ni(s)$

Here, nickel is getting reduced because it is gaining electrons and iron is getting oxidized because it is loosing electrons.

We know that:

$E^o_{(Fe^{3+}/Fe^{2+})}=0.77V\\E^o_{(Ni^{2+}/Ni)}=-0.23V$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

$E^o_{cell}=-0.23-0.77=-1.0V$

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$

As, the standard electrode potential of the cell is coming out to be negative for the above cell. Thus, the standard Gibbs free energy change of the reaction will become positive making the reaction non-spontaneous.

Hence, the above reaction is non-spontaneous.

3 0

2 years ago

CuSO4(aq) + 2NaOH(aq) Cu(OH)2(s) + Na2SO4(aq)

REY [17]

Your answer is right.

Important elements to consider:

- to use the balanced equation (which you did)
- divide the masses of each compound by the correspondant molar masses (which you did)
- compare the theoretical proportions with the current proportions

Theoretical: 2 mol of Na OH : 1 mol of CuSO4
Then 4 mol of NaOH need 2 mol of CUSO4.

Given that you have more than 2 mol of of CUSO4 you have plenty of it and the NaOH will consume first, being this the limiting reagent.

6 0

2 years ago

How many grams of sodium acetate ( molar mass = 83.06 g/mol ) must be added to 1.00 Liter of a 0.200 M acetic acid solution to m

Pie

<u>Answer:</u> The mass of sodium acetate that must be added is 30.23 grams

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of acetic acid solution = 0.200 M

Volume of solution = 1 L

Putting values in above equation, we get:

$0.200M=\frac{\text{Moles of acetic acid}}{1L}\\\\\text{Moles of acetic acid}=(0.200mol/L\times 1L)=0.200mol$

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[\text{salt}]}{[\text{acid}]})$

$pH=pK_a+\log(\frac{[CH_3COONa]}{[CH_3COOH]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of acetic acid = 4.74

$[CH_3COONa]=?mol$

$[CH_3COOH]=0.200mol$

pH = 5.00

Putting values in above equation, we get:

$5=4.74+\log(\frac{[CH_3COONa]}{0.200})$

$[CH_3COONa]=0.364mol$

To calculate the mass of sodium acetate for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of sodium acetate = 83.06 g/mol

Moles of sodium acetate = 0.364 moles

Putting values in above equation, we get:

$0.364mol=\frac{\text{Mass of sodium acetate}}{83.06g/mol}\\\\\text{Mass of sodium acetate}=(0.364mol\times 83.06g/mol)=30.23g$

Hence, the mass of sodium acetate that must be added is 30.23 grams

7 0

2 years ago

Which pair of compounds has the same empirical formula?

aleksandr82 [10.1K]

i think it's A. cause CH is 1:1 and if you reduce C2H2, the ratio would also be 1:1

5 0

2 years ago

Read 2 more answers