The reaction between sulfur dioxide and oxygen is reversible.

In the reaction of nitrogen gas with oxygen gas to produce nitrogen oxide, what is the effect of adding more oxygen gas to the i

Alborosie

<h3>Answer:</h3>

The Equilibrium would shift to produce more NO

<h3>Explanation:</h3>

The reaction is;

N₂(g) + O₂(g) ⇆ 2NO(g)

When a reaction is at equilibrium then the forward reaction rate will be equivalent to the reverse reaction rate. Additionally, the concentration of the reactants and products are the same.
From Le Chatelier's principle, additional reactants favor the formation of more products while additional products favor the formation of more reactants.
For example, when more oxygen is added then more Nitrogen (II) oxide will be formed.
Oxygen is a reactant and when increased it favors forward reaction which leads to the formation of more NO which is the product.

3 0

2 years ago

If 35.50 cm3 of a NaOH solution are required for the complete neutralization of a 25.00cm3 sample of 0.200mol dm-3 H2SO4, what i

Morgarella [4.7K]

In this question, you are given the NaOH volume but asked for concentration.
Don't forget that for every 1 mol of NaOH there will be 1 mol OH- ion, but for every 1 mol of H2SO4 there will be 2 mol of H- ion.
To neutralize you need the same amount of OH- and H+, so the equation should be:

OH-= H+
<span>35.50cm3 * x*1= 25cm3* 0.2mol/dm3 *2
</span>x= 10/35.5 mol/dm3= 0.2816/dm3

6 0

2 years ago

N2 and H2 are mixed in 14:3 mass ratio. After certain time ammonia was found to be 40% by mol. The mole fraction of N2 at that t

son4ous [18]

Answer : The mole fraction of nitrogen will be 0.4615.

Explanation : When nitrogen ( $N_{2}$ )and hydrogen ( $H_{2}$ )are mixed, the mole ratio becomes 1 : 1.5,

Now we know that ( $H_{2}$ ) is acting as a limiting agent.

So at the time of when 0.4 moles of ( $NH_{3}$ ) is been formed it requires 0.4 moles of ( $N_{2}$ ) and 3.4 moles of ( $H_{2}$ )

So, we find the the remaining ( $N_{2}$ ) will be 0.6 and

( $H_{2}$ ) will be 0.3 mole present in mixture.

So, the mole fraction of ( $N_{2}$ ) becomes = 0.6 / (0.6 + 0.4 + 0.3) Which becomes = 0.4615

8 0

2 years ago

Read 2 more answers

What is the formal charge on the nitrogen in hydroxylamine, h2noh?

andrew11 [14]

<h3>Answer:</h3>

Formal Charge on Nitrogen is "Zero".

<h3>Explanation:</h3>

Formal Charge on an atom in molecules is calculated using following formula;

Formal Charge = [# of Valence e⁻s] - [e⁻s in lone pairs + 1/2 # of Bonding e⁻s]

As shown in attached picture of Hydroxylamine, Nitrogen atom is containing two electrons in one lone pair of electrons and six electrons in three single bonds with two hydrogen and one oxygen atom respectively.

Hence,

Formal Charge = [5] - [2 + 6/2]

Formal Charge = [5] - [2 + 3]

Formal Charge = 5 - 5

Formal Charge = 0 (zero)

Hence, the formal charge on nitrogen atom in hydroxylamine is zero.