The difference samples A and B of Zinc oxide we're obtained from different sources when heated in a stream of hydrogen they were reduced to yield the following results
ZnO. Mass of sample. Mass of Zn red.
Sample. A 10.0g. 8.11g
Sample B 13.2g. 10.70
Show that the figures above illustrate the law of constant composition?
I am sure you meant "were" and not "we're".
Also I interpret your problem to mean mass Zn produced by reduction.
Sample A.
mass ZnO = 10.0
mass Zn = 8.11
mass O = 10.0 - 8.11 = 1.89. Then
mols Zn = 8.11/65.38 = 0.124
mols O = 1.89/16 = 0.118. Reduce these to a whole number to make it easier to compare.
Zn = 0.124/0.118 = 1.05
O = 0.118/0.118 = 1.00 and this ratio is essentially 1 to 1; hence the formula of ZnO.
I won't do the sample B part but it's done the same way and you should end up with similar ratio which proves the question.
Post your work if you get stuck.
Mass ZnO=13.2
Mass Zn=10.70
Mass O=13.2-10.70=2.5
Mols ZnO=10.70/65.38=0.1636
Mols O =2.5/16=0.156
Zn=0.1636/0.156=1.05
O=0.156/0.156=1.00
Am stuck I don't know how to do b
What is the formula for Zno
Well, it seems like the law of constant composition is at play here. Let me break it down for you with a touch of humor:
The law of constant composition states that a pure compound will always have the same elements in the same proportion by mass. Now, in this case, we have samples A and B of Zinc oxide.
Sample A, weighing in at a solid 10.0 grams, underwent the magical process of reduction by hydrogen, resulting in a transformation into Zn (red). After the transformation, it weighed 8.11 grams.
Sample B, on the other hand, had quite a big appetite, weighing a hefty 13.2 grams. It also experienced the hydrogen reduction extravaganza and ended up as Zn (red) with a weight loss down to 10.70 grams.
Now, if we examine the masses of Zn (red) obtained from both samples, we can see that Sample A lost 1.89 grams of oxygen, while Sample B said farewell to 2.50 grams of oxygen.
But here's where the law of constant composition shines through: despite the initial differences in mass, the ratio of zinc to oxygen in both samples remained constant! In other words, the proportion of zinc to oxygen in Zinc oxide is the same across different samples.
So, there you have it! The figures clearly illustrate the law of constant composition, proving that even in the mysterious world of chemistry, consistency and proportions matter. Keep it consistent, my friend!
To show that the figures above illustrate the law of constant composition, we need to compare the ratios of the masses of zinc and oxygen in each sample.
The law of constant composition states that a compound always contains the same elements in the same proportion by mass. In this case, zinc oxide (ZnO) is the compound we are analyzing.
Let's calculate the ratios for each sample:
Sample A:
Mass of Zinc (Zn) = 8.11g
Mass of Oxygen (O) = 10.0g - 8.11g = 1.89g
Ratio of Zn to O = 8.11g : 1.89g ≈ 4.29 : 1
Sample B:
Mass of Zinc (Zn) = 10.70g
Mass of Oxygen (O) = 13.2g - 10.70g = 2.50g
Ratio of Zn to O = 10.70g : 2.50g ≈ 4.28 : 1
Comparing these ratios, we can see that the ratios of zinc to oxygen in both samples A and B are approximately 4.3 : 1. This indicates that the relative composition of the elements in zinc oxide is consistent regardless of the source of the zinc oxide sample.
Therefore, based on the consistent ratios of zinc to oxygen, the figures above illustrate the law of constant composition.