Nitrogen dioxide gas is obtained by the reaction of nitrogen monoxide gas and oxygen gas . Write a balanced chemical equation fo
1 answer:
<u>Answer:</u> The equation is written below.
<u>Explanation:</u>
For the formation of nitrogen dioxide, the reactants used are nitrogen monoxide and oxygen gas.
The balanced chemical equation for the above reaction is given as:
By Stoichiometry of the reaction:
2 moles of nitrogen monoxide gas reacts with 1 mole of oxygen gas to produce 2 moles of nitrogen dioxide gas.
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The question is incomplete, here is the complete question:
Calculate the initial rate for the formation of C at 25°C, if [A]=0.50 M and [B]=0.075 M. Express your answer to two significant figures and include the appropriate units.Consider the reaction
A + 2B ⇔ C
whose rate at 25°C was measured using three different sets of initial concentrations as listed in the following table:
The table is attached below as an image.
<u>Answer:</u> The initial rate for the formation of C at 25°C is
<u>Explanation:</u>
Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.
For the given chemical equation:
Rate law expression for the reaction:
where,
a = order with respect to A
b = order with respect to B
- Expression for rate law for first trial:
....(1)
- Expression for rate law for second trial:
....(2)
- Expression for rate law for third trial:
....(3)
Dividing 2 by 1, we get:
Dividing 3 by 1, we get:
Thus, the rate law becomes:
......(4)
Now, calculating the value of 'k' by using any expression.
Putting values in equation 1, we get:
Calculating the initial rate of formation of C by using equation 4, we get:
[A] = 0.50 M
[B] = 0.075 M
Putting values in equation 4, we get:
Hence, the initial rate for the formation of C at 25°C is
The concentration of the solution reduces and the number of moles of solute isn't affected.
Data;
- V1 = 50mL
- C1 = 12.0M
- V2 = 200mL
- C2 = ?
1. When the solution is diluted, the concentration changes and this time, the concentration reduces.
Using dilution formula
The concentration of the solution reduces.
2. The number of moles remains the same.
When a solution is diluted, the number of moles remains the same because there's no change in the mass of the solute.
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