How many significant figures are there in the number 3.04

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

An object will sink in a liquid if the density of the object is greater than that of the liquid. the mass of a sphere is 9.83 g.

lord [1]

The volumeof this sphere must be less than 0.7228

5 0

2 years ago

When 9.2 g of frozen N2O4 is added to a 0.50 L reaction vessel and the vessel is heated to 400 K and allowed to come to equilibr

Amanda [17]

<u>Answer:</u> The value of $K_c$ for the given reaction is 1.435

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Given mass of $N_2O_4$ = 9.2 g

Molar mass of $N_2O_4$ = 92 g/mol

Volume of solution = 0.50 L

Putting values in above equation, we get:

$\text{Molarity of solution}=\frac{9.2g}{92g/mol\times 0.50L}\\\\\text{Molarity of solution}=0.20M$

For the given chemical equation:

$N_2O_4(g)\rightleftharpoons 2NO_2(g)$

<u>Initial:</u> 0.20

<u>At eqllm:</u> 0.20-x 2x

We are given:

Equilibrium concentration of $N_2O_4$ = 0.057

Evaluating the value of 'x'

$\Rightarrow (0.20-x)=0.057\\\\\Rightarrow x=0.143$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

$[NO_2]_{eq}=2x=(2\times 0.143)=0.286M$

$[N_2O_4]_{eq}=0.057M$

Putting values in above expression, we get:

$K_c=\frac{(0.286)^2}{0.143}\\\\K_c=1.435$

Hence, the value of $K_c$ for the given reaction is 1.435

6 0

2 years ago

How many grams of KClO3 are needed to produce of 4.26 moles of O2? 2 KClO3 2 KCl + 3 O2 a. 348 g b. 136 g c. 174 g d. 522 g e. 7

JulsSmile [24]

Ok so this is what we know :

2KClO3 -> 2KCl + 3O2 (Always check if equation is balanced - in this case it is)
4.26moles
So we know that we have 4.26 moles of oxygen (O2). Now lets look at the ratio between KClO3 and O2.
We see that the ratio is 2:3 meaning that we need 2KClO3 in order to produce 3O2.
Therefore divide 4.26 by 3 and then multiply by 2.
4.26/3 = 1.42
1.42 * 2 = 2.84
Now we know that the molarity of KClO3 is 2.84 moles.
Multiply by R.M.M to find how many grams of KClO3 we have.

R.M.M of KClO3
K- 39
Cl- 35.5
3O- 3 * 16 -> 48
---------------------------
<span>122.5
</span>2.84 * 122.5 = 347.9 grams therefore the answer is (a)
348 grams needed of KClO3 to produce 4.26 moles of O2.
Hope this helps :).

8 0

1 year ago

Read 2 more answers

Sodium hydroxide reacts with carbon dioxide to form sodium carbonate and water: 2 naoh(s) + co2(g) → na2co3(s) + h2o(l) how many

kow [346]

No of moles of naoh = 2.40 ÷ (23+16+1) = 0.06mol

no of moles of na2co3 = 0.06 ÷ 2 = 0.03mol

mass of na2co3 = 0.03 × (23×2+12+16×3) = 0.03 × 106 = 3.18g

8 0

1 year ago

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How many significant figures are there in the number 3.04​

How many significant figures are there in the number 3.04