An object weighting 9.6 grams is placed in a graduated cylinder displacing the volume from 10.0mL to 13.2 mL
1 answer:
The volume of object is 3.2 ml
<h3>Explanation:</h3>Given:
Mass of the object = M = 9.6 g
Initial volume of liquid:
Final volume of liquid after displacement:
Total volume of the displaced object inside a graduating cylinder will be given as difference between the final volume and initial volume of the expanding object.
V = 3.2 ml
You might be interested in
Answer:
Explanation:
Density can be found by dividing the mass by the volume.
The mass of the water is 50.0 grams.
The volume of the water is 50.0 milliliters.
Substitute the values into the formula.
Divide.
The density of the water is 1.0 grams per milliliter. Also, remember that the density of pure water is always 1.0 g/mL or g/cm³
<u>Answer:</u> NO is the limiting reagent in the given reaction and 0.857 moles of will be produced.
<u>Explanation:</u>
Limiting reagent is defined as the reagent which is present in less amount and it limits the formation of products.
Excess reagent is defined as the reagent which is present in large amount.
For the given chemical reaction:
We are given:
Moles of NO = 0.857 mol
Moles of oxygen = 0.498 mol
By stoichiometry of the reaction:
If 2 moles of NO reacts with 1 mole of oxygen gas.
So, 0.857 moles of NO will react with = of
As, the given amount of oxygen gas is more than the required amount. So, it is considered as an excess reagent.
Thus, NO is considered as the limiting reagent because it limits the formation of products.
By Stoichiometry of the reaction:
If 2 moles of NO produces 2 moles of nitrogen dioxide gas.
So, 0.857 moles of NO will produce = of
Hence, NO is the limiting reagent in the given reaction and 0.857 moles of will be produced.