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NeX [460]
3 years ago
14

An aqueous solution contains 0.397 M ammonia. Calculate the pH of the solution after the addition of 4.63 x 10-2 moles of perchl

oric acid (HClO4) to 250 mL of this solution. (Assume the volume does not change upon adding perchloric acid). Ka = 5.7 x 10-10, Kb = 1.80 x 10-5
Chemistry
1 answer:
SSSSS [86.1K]3 years ago
5 0

Answer:

9.308

Explanation:

The computation of the pH of the given solution is shown below:

But before we need to determine the HI molarity which is

Molarity\ of\ HI = \frac{moles}{volume \ in\ L}

= \frac{4.63\times10^{-2}}{0.250}

= 0.1852  M

Now

As we know that

NH_3 + HI = NH_4I

So,

NH_4I = HI = 0.1852 M

Now the molarity of NH_3 left is

= 0.397 - 0.1852

= 0.2118

pOH = pKb + log (\frac{NH_4I}{NH_3})

= 4.75 + log(\frac{0.1852}{0.2118})

= 4.692

Now as we know that

pH = 14 - pOH

= 14 - pOH

= 14 - 4.692

= 9.308

We simply applied the above equations

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Answer:

8.3 kJ

Explanation:

In this problem we have to consider that both water and the calorimeter absorb the heat of combustion, so we will calculate them:

q for water:

q H₂O = m x c x ΔT where m: mass of water = 944 mL x 1 g/mL = 944 g

                                      c: specific heat of water = 4.186 J/gºC

                                     ΔT : change in temperature = 2.06 ºC

so solving for q :

q H₂O = 944 g x 4.186 J/gºC x 2.06 ºC = 8,140 J

For calorimeter

q calorimeter  = C x  ΔT  where C: heat capacity of calorimeter = 69.6 ºC

                                     ΔT : change in temperature = 2.06 ºC

q calorimeter = 69.60J x 2.06 ºC = 143.4 J

Total heat released = 8,140 J +  143.4 J = 8,2836 J

Converting into kilojoules by dividing by 1000 we will have answered the question:

8,2836 J x 1 kJ/J = 8.3 kJ

7 0
3 years ago
Compute 6.28×1013+7.30×1011.
Simora [160]
6.28×1013+7.30×1011 this =13741.94
6 0
3 years ago
What is the function of a Mitochondrion​
gladu [14]

Answer:

Explanation:

Function. The mitochondrion is the site of ATP synthesis for the cell. The number of mitochondria found in a cell are therefore a good indicator of the cell's rate of metabolic activity; cells which are very metabolically active, such as hepatocytes, will have many mitochondria.

6 0
3 years ago
How long does it take to electroplate 0.5 mm of gold on an object with a surface area of 31 cm^^ from an Au3+(aq) solution with
konstantin123 [22]

Answer:

It will take 5492 seconds to electroplate 0.5 mm of gold on an object .

Explanation:

Mass of gold = m

Volume of gold = v

Surface area on which gold is plated = a=31 cm^2

Thickness of the gold plating  = h = 0.5 mm = 0.05 cm

1 mm = 0.1 cm

V=a\times h=31 cm^2\times 0.05 cm=1.55 cm^3

Density of the gold = d=19.3 g/cm^3

m=d\times v=19.3 g/cm^3\times 1.55 cm^3=29.915g

Moles of gold = \frac{29.915 g}{197 g/mol}=0.152 mol

Au^{3+}+3e^-\rightarrow Au

According to reaction, 1 mole of gold required 3 moles of electrons,then 0.152 moles of gold will require :

\frac{3}{1}\times 0.152 mol=0.456 mol of electrons

Number of electrons = N =0.456\times \times 6.022\times 10^{23}

Charge on single electron = q=1.6\times 10^{-19} C

Total charge required = Q

Q=N\times q

Amount of current passes = I = 8 Ampere

Duration of time  = T

I=\frac{Q}{T}

T=\frac{N\times q}{I}

=\frac{0.456\times \times 6.022\times 10^{23}\times 1.6\times 10^{-19} C}{8 A}=5492 s

It will take 5492 seconds to electroplate 0.5 mm of gold on an object .

7 0
3 years ago
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Natalka [10]
Organic elements, depends on the context of the question but I suspect that is the answer. The organic elements are H,O,C,N,P,S
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