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3 years ago

How are half-reaction useful?

Chemistry

2 answers:

Snezhnost [94]3 years ago

8 0

Answer:

Half-reactions are often used as a method of balancing redox reactions. For oxidation-reduction reactions in acidic conditions, after balancing the atoms and oxidation numbers, one will need to add H + ions to balance the hydrogen ions in the half reaction.

Explanation:

FrozenT [24]3 years ago

6 0

Answer:

We can use the half-reaction method to balance the equations of redox reactions occurring in aqueous solution. ... Each half-reaction is balanced for mass and charge, and then the two equations are recombined with appropriate coefficients so that the electrons cancel.

Explanation:

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How many moles of O2 should be supplied to burn 1.82 mol of C3H8 (propane) molecules In a camping stove?

pashok25 [27]

C3H8 + 5O2 ------> 3CO2 + 4H2O

from reaction 1 mol 5 mol

given 1.82 mol x mol

x=(1.82*5)/1 = 9.10 mol CO2

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3 years ago

Please help!!

prisoha [69]

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Then, what comes next is the word "react to form."
Finally, the products of the equation are mentioned.
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3 years ago

What is the noble gas electron configuration of bismuth (Bi)? A. [Kr] 6s2 6p3 B. [Xe] 6s2 6p3 C. [Xe] 4f14 5d10 6s2 6p3 D. [Kr]

OlgaM077 [116]

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3 years ago

Read 2 more answers

scandium47 has a half-life of 35s. suppose you have a 45g sample of scadium 47 how much of the sample remains unchanged after 14

Mars2501 [29]

The much of the sample that would remain unchanged after 140 seconds is 2.813 g

Explanation

Half life is time taken for the quantity to reduce to half its original value.

if the half life for Scandium is 35 sec, then the number of half life in 140 seconds

=140 sec/ 35 s = 4 half life

Therefore 45 g after first half life = 45 x1/2 =22.5 g

22.5 g after second half life = 22.5 x 1/2 =11.25 g

11.25 g after third half life = 11.25 x 1/2 = 5.625 g

5.625 after fourth half life = 5.625 x 1/2 = 2.813

therefore 2.813 g of Scandium 47 remains unchanged.

4 0

3 years ago

A 1.00 g sample of a metal X (that is known to form X ions in solution) was added to 127.9 mL of 0.5000 M sulfuric acid. After a

Semenov [28]

<u>Answer:</u> The metal having molar mass equal to 26.95 g/mol is Aluminium

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

Molarity of NaOH solution = 0.5000 M

Volume of solution = 0.03340 L

Putting values in equation 1, we get:

$0.5000M=\frac{\text{Moles of NaOH}}{0.03340L}\\\\\text{Moles of NaOH}=(0.5000mol/L\times 0.03340L)=0.01670mol$

The chemical equation for the reaction of NaOH and sulfuric acid follows:

$2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O$

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of sulfuric acid

So, 0.01670 moles of NaOH will react with = $\frac{1}{2}\times 0.01670=0.00835mol$ of sulfuric acid

Excess moles of sulfuric acid = 0.00835 moles

Calculating the moles of sulfuric acid by using equation 1, we get:

Molarity of sulfuric acid solution = 0.5000 M

Volume of solution = 127.9 mL = 0.1279 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.5000M=\frac{\text{Moles of }H_2SO_4}{0.1279L}\\\\\text{Moles of }H_2SO_4=(0.5000mol/L\times 0.1279L)=0.06395mol$

Number of moles of sulfuric acid reacted = 0.06395 - 0.00835 = 0.0556 moles

The chemical equation for the reaction of metal (forming $M^{3+}$ ion) and sulfuric acid follows:

$2X+3H_2SO_4\rightarrow X_2(SO_4)_3+3H_2$

By Stoichiometry of the reaction:

3 moles of sulfuric acid reacts with 2 moles of metal

So, 0.0556 moles of sulfuric acid will react with = $\frac{2}{3}\times 0.0556=0.0371mol$ of metal

To calculate the molar mass of metal for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Mass of metal = 1.00 g

Moles of metal = 0.0371 moles

Putting values in above equation, we get:

$0.0371mol=\frac{1.00g}{\text{Molar mass of metal}}\\\\\text{Molar mass of metal}=\frac{1.00g}{0.0371mol}=26.95g/mol$

Hence, the metal having molar mass equal to 26.95 g/mol is Aluminium

6 0

3 years ago