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Sedaia [141]
1 year ago
8

Use atomic properties to explain the reduction of a less active metal by a more active one:(b) in the molten state. Give a speci

fic example of each process.
Chemistry
1 answer:
Bad White [126]1 year ago
6 0

The reduction of a less active metal by a more active one is called metal displacement reactions. For example:

Fe + CuSO4 → FeSO4 + Cu

<h3>What is metal displacement reaction? </h3>

Displacement reactions is a reaction which includes a metal and the compound of a other metal. A more reactive metal will push or displace out a less reactive metal from its compound in this displacement reaction. The metal which is less reactive left uncombined after the reaction.

As we know that, electrons are the basis of the chemical reactions. If chemical compound or element A is more easily oxidized than B, then according to the terms of the activity series, the elements which are more easily oxidized can react with more chemicals, since they are able to act as a reducing agents for more chemicals.

Since, Metal ions are positively charged ions as they lose electrons. Some metals give up their electrons more readily than others and become more reactive.

Thus, we concluded that the reduction of a less active metal by a more active one is called metal displacement reactions. For example:

Fe + CuSO4 → FeSO4 + Cu

learn more about metal displacement reaction:

brainly.com/question/11777638

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How many molecules of oxygen are produced when a sample of 38.9 g of water is decomposed by electricity?
statuscvo [17]

Answer:

A) 6.5\times 10^{23}\ \text{molecules}

Explanation:

m = Mass of water = 38.9

M = Molar mass of water = 18 g/mol

N_A = Avogadro's number = 6.022\times 10^{23}\ \text{mol}^{-1}

The reaction of electrolysis would be

2H_2O(l)\rightarrow 2H_2(g)+O_2(g)

Number of moles of H_2O

n=\dfrac{m}{M}\\\Rightarrow n=\dfrac{38.9}{18}\ \text{mol}

From the reaction it can be seen that 2 moles of H_2O gives 1 mole of O_2

So, number of moles of O_2 produced is

\dfrac{38.9}{18}\times \dfrac{1}{2}=1.081\ \text{mol}

Number of molecules

1.081N_A=1.081\times 6.022\times 10^{23}\\ =6.5\times 10^{23}

So, 6.5\times 10^{23}\ \text{molecules} of oxygen is produced.

8 0
3 years ago
How do you calculate mass using density and volume?
skad [1K]
Example:

Mass = ?

Density = 25 g/mL

Volume = 5 mL

therefore:

d = m / V

25 = m / 5

m = 25 x 5

m = 125 g

hope this helps!

7 0
3 years ago
PLS HELP I NEED TO PASS THIS... In physical science lab, Kim and Dawn added small pieces of magnesium to hydrochloric acid. They
Bond [772]

B gas is produced is the answer.


8 0
3 years ago
A sample of pure lithium chloride contains 16% lithium by mass. What is the % lithium by mass in a sample of pure lithium carbon
Anna35 [415]

Answer:

Percentage lithium by mass in Lithium carbonate sample = 19.0%

Explanation:

Atomic mass of lithium = 7.0 g; atomic mass of Chlorine = 35.5 g; atomic mass of carbon = 12.0 g; atomic mass of oxygen = 16.0 g

Molar mass of lithium chloride, LiCl = 7 + 35.5 = 42.5 g

Percentage by mass of lithium in LiCl = (7/42.5) * 100% = 16.4 % aproximately 16%

Molar mass of lithium carbonate, Li₂CO₃ = 7 * 2 + 12 + 16 * 3 =74.0 g

Percentage by mass of lithium in Li₂CO₃ = (14/74) * 100% = 18.9 % approximately 19%

Mass of Lithium carbonate sample = 2 * 42.5 = 85.0 g

mass of lithium in 85.0 g Li₂CO₃ = 19% * 85.0 g = 16.15 g

Percentage by mass of lithium in 85.0 g Li₂CO₃ = (16.15/85.0) * 100 % = 19.0%

Percentage lithium by mass in Lithium carbonate sample = 19.0%

3 0
3 years ago
Calculate the amount of energy required to melt 500 g of ice at 0oc. (δhfus=5.96 kj/mole; δhfus is the energy required to melt i
marta [7]
When you want to melt an ice, you only need the latent energy of fusion, <span>δhfus. We use the given value, then multiply this with the given amount to determine the amount of energy. Since the energy is per mole basis, use the molar mass of ice which is 18 g/mol. The solution is as follows:

</span>ΔH = 5.96 kJ/mol * 1 mol/18 g * 500 g
<em>ΔH = 165.56 kJ</em><span>
</span>
3 0
3 years ago
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