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Schach [20]
3 years ago
12

WHAT ARE NEWTON'S 3 LAW'S

Chemistry
1 answer:
Tanzania [10]3 years ago
6 0

1. An object at rest will remain at rest unless acted on by an unbalanced force. An object in motion continues in motion with the same speed and in the same direction unless acted upon by an unbalanced force. Also know as “the law of inertia"

2. Acceleration is produced when a force acts on a mass. The greater the mass (of the object being accelerated) the greater the amount of force needed (to accelerate the object).

3. For every action there is an equal and opposite re-action.
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In the first 15.0 s of the reaction, 1.9×10−2 mol of O2 is produced in a reaction vessel with a volume of 0.480 L . What is the
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The question is incomplete, here is the complete question:

Consider the following reaction:  2N_2O(g)\rightarrow 2N_2(g)+O_2(g)

In the first 15.0 s of the reaction, 1.9×10⁻² mol of O₂ is produced in a reaction vessel with a volume of 0.480 L . What is the average rate of the reaction over this time interval?

<u>Answer:</u> The average rate of appearance of oxygen gas is 2.64\times 10^{-3}M/s

<u>Explanation:</u>

We are given:

Moles of oxygen gas = 1.9\times 10^{-2}moles

Volume of solution = 0.480 L

Molarity is calculated by using the equation:

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

So, \text{Molarity of }O_2=\frac{1.9\times 10^{-2}mol}{0.480L}=0.0396M

The given chemical reaction follows:

2N_2O(g)\rightarrow 2N_2(g)+O_2(g)

The average rate of the reaction for appearance of O_2 is given as:

\text{Average rate of appearance of }O_2=\frac{\Delta [O_2]}{\Delta t}

Or,

\text{Average rate of appearance of }O_2=\frac{C_2-C_1}{t_2-t_1}

where,

C_2 = final concentration of oxygen gas = 0.0396 M

C_1 = initial concentration of oxygen gas = 0 M

t_2 = final time = 15.0 s

t_1 = initial time = 0 s

Putting values in above equation, we get:

\text{Average rate of appearance of }O_2=\frac{0.0396-0}{15-0}\\\\\text{Average rate of appearance of }O_2=2.64\times 10^{-3}M/s

Hence, the average rate of appearance of oxygen gas is 2.64\times 10^{-3}M/s

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