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Fynjy0 [20]
1 year ago
12

If 0. 5371 g of zn was plated onto an electrode, how much charge (in coulombs) was used to reduce the zn2+ to zn(s)?

Chemistry
1 answer:
My name is Ann [436]1 year ago
7 0

The amount of charge required to reduce Zn²⁺ to Zn is 1585.26 coulomb.

<h3>What is electrolysis?</h3>

The process that has been equipped with the use of electric current to break the chemical compound and deposited at the constituent electrodes is termed as electrolysis.

The reaction of reduction of Zn can be given as:

Zn²⁺ + 2e⁻ → Zn (s)

Molar mass of Zn = 65.39 g

The equivalent of Zn²⁺ has been:

Equivalent = atomic mass/equivalent weight

Equivalent Zn²⁺ = 65.39/32.69

Equivalent Zn²⁺ = 2 mol

1 equivalent = 1 F charge used

2 mol equivalent = 2 F charge used

Charge in coulomb can be given as:

1 Faraday = 96500 coulomb

2 F = 96500 * 2 coulomb

2 F = 1,93,000 coulomb

The moles of Zn to be deposited is:

Moles = weight /molar weight

Moles = 0.5371 g/ 65.39 g/mol

Moles = 0.00821 mol

Therefore, the charge used for the deposition of 0.00821 mol is:

Charge = 0.00821 * 1,93,000 coulomb

Charge = 1585.26 coulomb

1585.26 coulomb charge was used to reduce the Zn²⁺ to Zn.

Learn more about electrolysis, Here:

brainly.com/question/12994141

#SPJ4

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nikitadnepr [17]

Answer: m = n·M = 34.7 g

Explanation:

M(Li) = 6.941 g/mol, n = 5 mol

6 0
3 years ago
Pure chlorobenzene (C6H5Cl) has a normal boiling point of 131.00 °C. A solution of 32.5 g of 2,8-dibromodibenzofuran (C12H6Br2O)
vichka [17]

Answer:

Kb →  1.56 °C / m

Explanation:

This is all about boiling point elevation, the colligative property that shows that boiling point for a solution is higher than boiling point of pure solvent.

This is the formula: ΔT = Kb . m . i

where i is the Van't Hoff factor (ions dissolved in solution). As these are organic compounds, we assume they are non electrolytic,

m is molality (mol of solute / 1kg of solvent)

Kb is our unknown. The value for ebulloscopic constant, it is specific for each solvent.

ΔT = T° boiling from solution - T° boiling from solute

First of all, let's determine the moles of solute.

Mass / Molar mass → 32.5 g/ 113.45 g/mol = 0.286 mol

Molality is mol of solute/ 1 kg of solvent

We must convert the mass from g to kg

195g . 1kg /1000 = 0.195 kg

Molality = 0.286 mol / 0.195 kg = 1.47 m

Let's replace the values in the formula

133.30 °C - 131°C = Kb . 1.47m .1

2.30°C / 1.47 m =  Kb →  1.56 °C / m

3 0
3 years ago
Using the data, which of the following is the rate constant for the rearrangement of methyl isonitrile at 320 °C? (HINT: the act
algol13

Answer:

D) 2.3 x 10⁻¹ s⁻¹

Explanation:

The rate constant is related to the activation energy through the formula:

k= Ae^(-Eₐ /RT)

where A is the collision factor, Eₐ the activation energy, R is the gas constant ( 8.314 J/Kmol ) , and T is the temperature (K)

So a plot of lnk versus 1/T ( Arrehenius plot ) gives us a straight line with slope equal -Eₐ/R and intercept lnA

lnk = -(Eₐ/T)(1/T) + lnA

which has the form y= mx + b

In this problem, we can use the data provided to:

a) Using a calculator determine the slope and intercept and then calculate the value of rate constant at 320 ºC, or

b) Plot the data and determine the equation of the best line , and answer the question for k @ 320 ºC by reading the value from the plot.

Once you do the plot, the resulting equation is:

y = - 19 x 10³ x + 30,582 ( R² = 0.999 )

So for T = 320 + 273 K = 593 K

Y = 19 x 10³ X + 30.58

So for T = (320 + 273)K = 593 K

Y =  -19 x 10³ ( 1/593) + 30.58 = -32.04 +30.58 = - 1.46

and then since

y = lnk ⇒ e^y = k

k= e^-1.46 = 2.3 x 10⁻¹ s⁻¹

Note: there is an error of transcription in the value for T = 472.1 ( 1/T = 2.118 x 10⁻³  and  not 2.228 x 10⁻³). You can  recognize this mistake if you plot the data and notice it produces an outlier.

5 0
3 years ago
The nucleus of an atom stays together only because the repulsive forces, called
Ahat [919]

Answer:

Electrostatic

Explanation:

The forces that are overcome are the repulsive electrostatic forces between the protons (all charged positively).

7 0
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To standardize a hydrochloric acid solution, it was used as a titrant with a solid sample of sodium hydrogen carbonate, NaHCO3.
Ket [755]

Answer:

0.113 M

Explanation:

The reaction that takes place is:

  • NaHCO₃ + HCl →NaCl + CO₂ + H₂O

First we convert 0.3967 g of NaHCO₃ into moles, using its molar mass:

  • 0.3967 g ÷ 84 g/mol = 4.72x10⁻³ mol NaHCO₃

As 1 mol of NaHCO₃ reacts with 1 mol of HCl, in 41.77 mL of the HCl solution there were 4.72x10⁻³ moles of HCl.

With the <em>calculated number of moles and the given volume </em>we <u>calculate the concentration of the solution</u>:

  • Converting 41.77 mL ⇒ 41.77 mL / 1000 = 0.04177 L
  • Concentration = 4.72x10⁻³ mol / 0.04177 L = 0.113 M
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