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2 years ago

Write a balanced half-reaction describing the oxidation of gaseous dihydrogen to aqueous hydrogen cations.

2 answers:

8 0

The oxidation state of hydrogen gas is 0 and oxidation state of hydrogen cation is +1.
There’s an increase in oxidation number therefore it’s an oxidation reaction.
Oxidation reactions give out electrons. The masses and charges on both sides should be balanced
Half reaction is
H2 —> 2H+ +2e

Vilka [71]2 years ago

8 0

Answer: $H_2(g\rightarrow 2H^+(aq)+2e^-$

Explanation:

Oxidation-reduction reaction or redox reaction is defined as the reaction in which oxidation and reduction reactions occur simultaneously.

Oxidation reaction is defined as the reaction in which a substance looses its electrons. The oxidation state of the substance increases.

Reduction reaction is defined as the reaction in which a substance gains electrons. The oxidation state of the substance gets reduced.

$H_2(g)\rightarrow 2H^+(aq)+2e^-$

On reactant side:

Oxidation state of hydrogen = 0

On product side:

Oxidation state of $H^+$ = +1

The oxidation state of hydrogen increases from 0 to +1. Thus, it is getting oxidized and it undergoes oxidation reaction.

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If you gently shake a carbon dioxide fire extinguisher, you will feel the presence of liquid within the extinguisher What condit

scoundrel [369]

When we wish to convert a gas to liquid we have to either

a) decrease temperature

b) increase pressure

In case of fire extinguisher the CO2 is found to be in liquid state, this is as the CO2 is pressurized at high pressure which keeps CO2 in liquid state

the ideal pressure and temperature conditions when CO2 gas can be converted to CO2 gas

Pressure = 5 - 73 atm

Temperature = -57 to 31 degree Celsius

6 0

2 years ago

Which of the following represents a propagation step in the monochlorination of methylene chloride (CH2Cl2)?a. CHCl3 + Cl. Right

svetlana [45]

Answer:

B = CHCl2 + Cl2 --> CHCl3 + Cl

Explanation:

Free radical halogenation is a chlorination reaction on Alkane hydrocarbons. This involves the splitting of molecules into radicals/ unstable molecules in the presence of sunlight/ U.V light which ensures bonding of the molecules.

Free radical chlorination is divided into 3 steps which are:

The initiation step

The propagation step

The termination step

So in reference to the question, propagation step involves two steps.

The first step is where the molecule in this case the methylene chloride(CH2Cl2) loses a hydrogen atom and then bond with a chlorine atom radical to give a nethylwnw chloride radical and HCl.

The second step involves the reaction of this methylene chloride got in the first step with chlorine molecule to form trichloride methane and a chlorine radical.

You would find in the attachment the 2 step mechanism.

3 0

2 years ago

A large cylindrical tank contains 0.750 m3 of nitrogen gas at 27°C and 7.50 * 103 Pa (absolute pressure). The tank has a tight‐f

Nuetrik [128]

Answer: The answer is 68142.4 Pa

Explanation:

Given that the initial properties of the cylindrical tank are :

Volume V1= 0.750m3

Temperature T1= 27C

Pressure P1 =7.5*10^3 Pa= 7500Pa

Final properties of the tank after decrease in volume and increase in temperature :

Volume V2 =0.480m3

Temperature T2 = 157C

Pressure P2 =?

Applying the gas law equation (Charles and Boyle's laws combined)

P1V1/T1 = P2V2/T2

(7500 * 0.750)/27 =( P2 * 0.480)/157

P2 =(7500 * 0.750* 157) / (0.480 *27)

P2 = 883125/12.96

P2 = 68142.4Pa

Therefore the pressure of the cylindrical tank after decrease in volume and increase in temperature is 68142.4Pa

8 0

2 years ago

Website

vredina [299]

Answer:

volume in L = 0.25 L

Explanation:

Given data:

Mass of Cu(NO₃)₂ = 2.43 g

Volume of KI = ?

Solution:

Balanced chemical equation:

2Cu(NO₃)₂ + 4KI → 2CuI + I₂ + 4KNO₃

Moles of Cu(NO₃)₂:

Number of moles = mass/ molar mass

Number of moles = 2.43 g/ 187.56 g/mol

Number of moles = 0.013 mol

Now we will compare the moles of Cu(NO₃)₂ with KI.

Cu(NO₃)₂ : KI

2 : 4

0.013 : 4 × 0.013=0.052 mol

Volume of KI:

Molarity = moles of solute / volume in L

volume in L = moles of solute /Molarity

volume in L = 0.052 mol / 0.209 mol/L

volume in L = 0.25 L

6 0

2 years ago

Albus Dumbledore provides his students with a sample of 19.3 g of sodium sulfate. How many oxygen atoms are in this sample

Dimas [21]

Answer:

3.27·10²³ atoms of O

Explanation:

To figure out the amount of oxygen atoms in this sample, we must first evaluate the sample.

The chemical formula for sodium sulfate is Na₂SO₄, and its molar mass is approximately 142.05 $\frac{g}{mol}$ .

We will use stoichiometry to convert from our mass of Na₂SO₄ to moles of Na₂SO₄, and then from moles of Na₂SO₄ to moles of O using the mole ratio; then finally, we will convert from moles of O to atoms of O using Avogadro's constant.

19.3g Na₂SO₄ · $\frac{1 mol Na^2SO^4}{142.05g Na^2SO^4}$ · $\frac{4 mol O}{1 mol Na^2SO^4}$ · $\frac{6.022x10^2^3}{1 mol O}$

After doing the math for this dimensional analysis, you should get a quantity of approximately 3.27·10²³ atoms of O.

3 0

2 years ago