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2 years ago

How many Faradays of electricity are carried by 1.505×10^23 electronsk

Chemistry

1 answer:

aksik [14]2 years ago

3 0

Mole of electron required by $1.505 *10^{23}$ mole is $2.5* 10 ^{-1}$

Faraday law expressed how the change that is been being produced by a current at an electrode-electrolyte interface is related and proportional to the quantity of electricity that is been used.

There is one mole of electron required for 1 Faraday of electricity.

Avogadro constant is $6.02*10^{23}$

Mole of electron can be calculated by dividing the number of electron by avogadro's constant.

= $\frac{1.505*10^{23} }{6.02*10^{23} }$

= $2.5* 10^{-1} Faraday of electricity$

Therefore, it requires $2.5*10^{-1}$ Faraday of electricity for the $1.505 * 10^{23} mole.$

Learn more at: brainly.com/question/1640558?referrer=searchResults

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Determine the change in boiling point for 397.7 g of carbon disulfide (Kb = 2.34°C kg/mol) if 35.0 g of a nonvolatile, nonionizi

uysha [10]

Answer: The change in boiling point for 397.7 g of carbon disulfide (Kb = 2.34°C kg/mol) if 35.0 g of a nonvolatile, nonionizing compound is dissolved in it is $2.9^0C$

Explanation:

Elevation in boiling point:

$T_b-T^o_b=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_b$ = boiling point of solution = ?

$T^o_b$ = boiling point of pure carbon disulfide= $46.2^oC$

$k_b$ = boiling point constant = $2.34^0Ckg/mol$

m = molality

i = Van't Hoff factor = 1 (for non-electrolyte)

$w_2$ = mass of solute = 35.0 g

$w_1$ = mass of solvent (carbon disulphide) = 397.7 g

$M_2$ = molar mass of solute = 70.0 g/mol

Now put all the given values in the above formula, we get:

$(T_b-46.2)^oC=1\times (2.34^oC/m)\times \frac{(35.0g)\times 1000}{70.0\times (397.7g)}$

$T_b=49.1^0C$

Therefore, the change in boiling point is $(49.1-46.2)^oC=2.9^0C$

5 0

3 years ago

State the oxidation number of S in <br><img src="https://tex.z-dn.net/?f=H_%7B2%7DSO_%7B3%7D" id="TexFormula1" title="H_{2}SO_{3

Nezavi [6.7K]

Taking into account the definition of oxidation number, the oxidation numbers of S in H₂SO₃ is 4.

<h3>Definition of oxidation number</h3>

The oxidation number is the charge that an atom has; is an integer that represents the number of electrons an atom puts into play when it forms a given compound.

In other words, the oxidation number of an element is a value that indicates the number of electrons that element gains or loses when it combines with another.

<h3>Oxidation number determination</h3>

To determine the oxidation state of different elements it is necessary to know that:

The oxidation number of hydrogen in a compound is +1, except in metal hydrides, where is –1.
The oxidation number of oxygen in a compound is –2, except in peroxides, where it is –1.

On the other side, the sum of the oxidation numbers of the existing elements in a chemical formula must add up to zero.

Then, considering the oxidation numbers of each element, multiplying it by the number of existing elements in the chemical formula and adding it and equaling it to zero, the value of the missing oxidation number can be obtained.

<h3>Oxidation numbers of S</h3>

In this case, the oxidation numbers of S in H₂SO₃ is calculated as:

2× (+1) + oxidation numbers of S + 3×(-2)= 0

2 + oxidation numbers of S -6= 0

oxidation numbers of S -4= 0

<u><em>oxidation numbers of S= 4</em></u>

Finally, the oxidation numbers of S in H₂SO₃ is 4.

Learn more about the oxidation number:

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5 0

2 years ago

Read 2 more answers

If 12.1 kilograms of al2o3(s), 60.4 kilograms of naoh(l), and 60.4 kilograms of hf(g) react completely, how many kilograms of cr

Fynjy0 [20]

Answer is: 7,826 kg of cryolite.
Chemical reaction: Al₂O₃ + 6NaOH + 12HF → 2Na₃AlF₆ + 9H₂<span>O.
m(</span>Al₂O₃) = 12,1 kg = 12100 g.
n(Al₂O₃) = m(Al₂O₃) ÷ M(Al₂O₃).
n(Al₂O₃) = 12100 g ÷ 101,96 g/mol = 111,86 mol; limiting reactant.
m(NaOH) = 60,4 kg = 60400 g.
n(NaOH) = 60400 g ÷ 40 g/mol.
n(NaOH) = 1510 mol.
m(HF) = 60,4 kg = 60400 g.
n(HF) = 60400 g ÷ 20 g/mol = 3020 mol.
From chemical reaction: n(Al₂O₃) : n(Na₃AlF₆) = 6 : 2.
n(Na₃AlF₆) = 2 ·111,86 mol ÷ 6 = 37,28 mol.
m(Na₃AlF₆) = 37,28 mol · 209,94 g/mol.
m(Na₃AlF₆) = 7826,56 g = 7,826 kg.

7 0

3 years ago

Atomic weight of lithium ?

LuckyWell [14K]

Answer:

6.941

Explanation:

5 0

3 years ago

Please help ASAP<br> Will give brainliest

Leni [432]

Answer: 457.8

Explanation:

345.8*225/703.55

8 0

2 years ago

How many Faradays of electricity are carried by 1.505×10^23 electronsk​

How many Faradays of electricity are carried by 1.505×10^23 electronsk