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sergiy2304 [10]
3 years ago
12

An electrode has a potential of 1.201 V with respect to a saturated silver-silver chloride electrode. What would the electrodes

potential be relative to a saturated calomel electrode (S.C.E.)? E = 0.197 V for the saturated silver-silver chloride electrode and E = 0.241 V for the S.C.E.
Chemistry
1 answer:
fredd [130]3 years ago
7 0

Answer:

The potential wrt. calomel is 1.254 V

Explanation:

Given:

Potential wrt. silver chloride E_{Ag}  = 1.201 V

Potential wrt. saturated silver chloride E = 0.197 V

Potential wrt. SCE E_{Hg} = 0.241 V

Now potential wrt. hydrogen is given by,

   = 1.201- 0.197

   = 1.004 V

And we find for potential wrt. calomel,

   = potential wrt. hydrogen + potential wrt. SEC

   = 1.004 +0.241

   = 1.254 V

Therefore, the potential wrt. calomel is 1.254 V

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Alenkasestr [34]

<u>Answer:</u> The sample of Cobalt-60 isotope must be replaced in January 2027

<u>Explanation:</u>

The equation used to calculate rate constant from given half life for first order kinetics:

t_{1/2}=\frac{0.693}{k}

where,

t_{1/2} = half life of the reaction = 5.26 years

Putting values in above equation, we get:

k=\frac{0.693}{5.26yrs}=0.132yrs^{-1}

Rate law expression for first order kinetics is given by the equation:

k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}

where,

k = rate constant = 0.132yr^{-1}

t = time taken for decay process = ? yr

[A_o] = initial amount of the sample = 100 grams

[A] = amount left after decay process =  (100 - 75) = 25 grams

Putting values in above equation, we get:

0.132=\frac{2.303}{t}\log\frac{100}{25}\\\\t=10.5yrs

The original sample was purchased in June 2016

As, June is the 6th month of the year, which means the time period will be 2016+\frac{6}{12}=2016.5

Adding the time in the original time period, we get:

2016.5+10.5=2027

Hence, the sample of Cobalt-60 isotope must be replaced in January 2027

3 0
4 years ago
Fran hangs three balloons from strings. She gives the first balloon a positive charge. She gives the second balloon a negative c
fgiga [73]

Explanation: We are given three balloon carrying different charges: One having positive charge, one having negative charge and one having no charge.

When we bring positive rod near these three balloons, all of them behave differently.

According to the Coulomb's Law:

Like charges repel each other and unlike charges attract each other.

  • When we bring positive rod towards the balloon having positive charge, the balloon will move away from the rod, because like charges repel each other.
  • When we bring positive rod near negative balloon, the balloon will be attracted to the rod because unlike charges attract each other.
  • But, when we bring positive rod towards the balloon having no charge, the charges will be induced in the balloon. Although the balloon is neutral in nature but it will still carry some +ve or -ve charges. So, when the positive rod is bought near neutral ball, the negative side of the balloon will get attracted towards the rod and positive side will be repelled. From this, we say that the rod has induced some charges in the balloon.
5 0
3 years ago
A pupil adds 5 cm of 12 mol.dm sulphuric acids to make a 250 cmº solution. Calculate the
klemol [59]

Answer:

0.500 mol/dm³

Explanation:

Using the formula below;

CaVa = CbVb

Where;

Ca = concentration of acid (mol/dm³)

Cb = concentration of base (mol/dm³)

Va = volume of acid (cm³)

Vb = volume of base (cm³)

In accordance to the information provided in this question is;

Va = 5cm³

Vb = 250 cm³

Ca = 12 mol/dm³

Cb = ?

Using CaVa = CbVb

12 × 5 = Cb × 250

60 = 120Cb

Cb = 60/120

Cb = 0.500 mol/dm³

8 0
3 years ago
When a solution of 0.1 M Mg(NO3)2 was mixed with a limited amount of aqueous ammonia, a light white, wispy solid was observed, i
Ipatiy [6.2K]

<u>Answer:</u> The net ionic equation is written below.

<u>Explanation:</u>

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.

The chemical equation for the reaction of magnesium nitrate and aqueous ammonia (ammonium hydroxide) is given as:

Mg(NO_3)_2(aq.)+2NH_4OH(aq.)\rightarrow Mg(OH)_2(s)+2NH_4NO_3(aq.)

A white precipitate of magnesium hydroxide is formed in the above reaction.

Ionic form of the above equation follows:

Mg^{2+}(aq.)+2NO_3^-(aq.)+2NH_4^+(aq.)+2OH^-(aq.)\rightarrow Mg(OH)_2(s)+2NH_4^+(aq.)+2NO_3^-(aq.)

As, ammonium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

Mg^{2+}(aq.)+2OH^-(aq.)\rightarrow Mg(OH)_2(s)

Hence, the net ionic equation is written above.

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telo118 [61]

Answer:

The pH of the sweater containing Hydrogen ion concentration

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<u>8</u>

<u></u>

Explanation:

pH = It is the negative logarithm of activity (concentration) of hydrogen ions.

pH = -log([H+])

Now, In the question the concentration of [H+] ions is :

[H^{+}]=1\times 10^{-8}

pH = -log(1\times 10^{-8})

use the relation:

log(10^{-a})=a

pH= -(-8)

pH = 8

Note : <em><u> 1 times 10 to the power of 8 must be" 1 times 10 to the power of -8"</u></em>

If the concentration is

[H^{+}]=1\times 10^{8}

Then pH = -8 , which is not possible . So in that  case the pH calculation is by other method

5 0
3 years ago
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