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Schach [20]
1 year ago
14

How many amps are required to produce 75. 8 g of iron metal from a solution of aqueous iron(iii)chloride in 6. 75 hours?

Chemistry
1 answer:
Shalnov [3]1 year ago
7 0

The amount of current required to produce 75. 8 g of iron metal from a solution of aqueous iron (iii)chloride in 6. 75 hours is 168.4A.

The amount of Current required to deposit a metal can be find out by using The Law of Equivalence. It states that the number of gram equivalents of each reactant and product is equal in a given reaction.

It can be found using the formula,

m = Z I t

where, m = mass of metal deposited = 75.8g

            Z = Equivalent mass / 96500 = 18.6 / 96500 = 0.0001

             I is the current passed

              t is the time taken = 75hour = 75 × 60 = 4500s

On subsituting in above formula,

75.8 = E I t / F

⇒ 75.8 = 0.0001 × I × 4500

⇒ I = 168.4 Ampere (A)

Hence, amount of current required to deposit a metal is 168.4A.

Learn more about Law of Equivalence here, brainly.com/question/13104984

#SPJ4

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What is the molar mass of an element?
Vesnalui [34]

The answer is: the mass of 6.02 x 1023 representative particles of the element.

The base SI unit for molar mass is kg/mol, but chemist more use g/mol (gram per mole).

For example, molar mas of ammonia is 17.031 g/mol.

M(NH₃) = Ar(N) + 3 · Ar(H) · g/mol.

M(NH₃) = 14.007 + 3 · 1.008 · g/mol.

M(NH₃) = 17.031 g/mol.

The molar mass (M) is the mass of a given substance (in this example ammonia) divided by the amount of substance.


8 0
3 years ago
How many liters of hydrogen are required to react completely with 2.4L of oxygen to form water? 2H2 + O2 --> 2H2O
RoseWind [281]

Answer:

2.4 mole of oxygen will react with 2.4 moles of hydrogen

Explanation:

As we know

1 liter = 1000 grams

2H2 + O2 --> 2H2O

Weight of H2 molecule = 2.016 g/mol

Weight of water = 18.01 gram /l

2 mole of oxygen react with 2 mole of H2

2.4 mole of oxygen will react with 2.4 moles of hydrogen

3 0
2 years ago
How many moles of CaCO3 are needed to react with 12.5 mol SO2​
grandymaker [24]
<h3><u>Answer;</u></h3>

= 12.5 Moles of CaSO3

<h3><u>Explanation</u>;</h3>

The reaction between CaCO3 and SO2 is given by the equation.

CaCO3(s) + SO2(g) → CaSO3(aq) + CO2(g)

The mole ratio between CaCO3 and SO2 is 1 : 1;

1 mole of CaCO3 reacts with 1 mole SO2 to form CaSO3 and CO2

Therefore;

<em>12.5 moles of SO2 will require 12.5 moles of CaSO3</em>

7 0
3 years ago
The Henry's law constant (kH) for O2 in water at 20°C is 1.28 × 10−3 mol/(L·atm). (a) How many grams of O2 will dissolve in 4.00
Burka [1]

Answer:

Solubility of O₂(g) in 4L water = 3.42 x 10⁻² grams O₂(g)

Explanation:

Graham's Law => Solubility(S) ∝ Applied Pressure(P) => S =k·P

Given P = 0.209Atm & k = 1.28 x 10⁻³mol/L·Atm

=> S = k·P = (1.28 x 10⁻³ mole/L·Atm)0.209Atm = 2.68 x 10⁻³ mol O₂/L water.

∴Solubility of O₂(g) in 4L water at 0.209Atm = (2.68 x 10⁻³mole O₂(g)/L)(4L)(32 g O₂(g)/mol O₂(g)) = <u>3.45 x 10⁻² grams O₂(g) in 4L water. </u>

4 0
2 years ago
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