<h3>Answers:</h3>
1) 2 Units of Ozone
2) 3 Units of Ozone
3) 9 Units of Ozone
<h3>Solution:</h3>
1) From 6 Oxygen Particles;
As given,
3 Oxygen Particles form = 1 Unit of Ozone
So,
6 Oxygen Particles will form = X Units of Ozone
Solving for X,
X = (6 O Particles × 1 Unit of Ozone) ÷ 3 O Particles
X = 2 Units of Ozone
2) From 9 Oxygen Particles;
As given,
3 Oxygen Particles form = 1 Unit of Ozone
So,
9 Oxygen Particles will form = X Units of Ozone
Solving for X,
X = (9 O Particles × 1 Unit of Ozone) ÷ 3 O Particles
X = 3 Units of Ozone
3) From 27 Oxygen Particles;
As given,
3 Oxygen Particles form = 1 Unit of Ozone
So,
27 Oxygen Particles will form = X Units of Ozone
Solving for X,
X = (27 O Particles × 1 Unit of Ozone) ÷ 3 O Particles
X = 9 Units of Ozone

Here's the balanced equation for given Double displacement reaction ~

The products fored are : Lead Iodide ( PbI2 ) and Potassium Nitrate ( KNO3 )
Enter an equation of a chemical reaction and click 'Balance'. The answer will appear below
Always use the upper case for the first character in the element name and the lower case for the second character. Examples: Fe, Au, Co, Br, C, O, N, F. Compare: Co - cobalt and CO - carbon monoxide
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Example: Fe{3+} + I{-} = Fe{2+} + I2
Substitute immutable groups in chemical compounds to avoid ambiguity.
For instance equation C6H5C2H5 + O2 = C6H5OH + CO2 + H2O will not be balanced,
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Compound states [like (s) (aq) or (g)] are not required.
If you do not know what products are enter reagents only and click 'Balance'. In many cases a complete equation will be suggested.
Answer:
B. Measures of central tendency
Explanation:
Mean, median and mode are best described as measures of central tendency of a given data set.
Mean is the average of the samples given
Mode is the data point with the most frequent occurrence
Median is the data point that lies in the middle
- All these parameters tells us how far a data point is from the middle or how close they are.
Answer : The correct option is (d) 1.5 g K, 0.38 g O₂
Explanation :
The molar masses of potassium and oxygen are very close to each other. Therefore we can assume them to be equal. If we assume that, then according to reaction stoichiometry, 4 moles of K are needed to react with 1 mol of O₂. Since the molar masses are assumed to be equal , we can say that the mass of potassium needed to react with that of oxygen should be 4 times the mass of oxygen.
From the given options, the only option that has less amount of K is option d.
Here, 0.38 g of oxygen needs 0.38 x 4 = 1.52 g of K. But the given mass of potassium is 1.5 g which is less. This indicates that potassium is the limiting reactant as we do not have enough potassium to completely react with all of the oxygen.
Therefore option d is the correct option