there are 6 moles of oxygen in Zn(No3)2
Answer:
<em>3.27·10²³ atoms of O</em>
Explanation:
To figure out the amount of oxygen atoms in this sample, we must first evaluate the sample.
The chemical formula for sodium sulfate is <em>Na₂SO₄, </em>and its molar mass is approximately 142.05
.
We will use stoichiometry to convert from our mass of <em>Na₂SO₄ </em>to moles of <em>Na₂SO₄</em>, and then from moles of <em>Na₂SO₄ </em>to moles of <em>O </em>using the mole ratio; then finally, we will convert from moles of <em>O </em>to atoms of <em>O </em>using Avogadro's constant.
19.3g <em>Na₂SO₄</em> · 
· 
·
After doing the math for this dimensional analysis, you should get a quantity of approximately <em>3.27·10²³ atoms of O</em>.
Similar chemical properties
(OPTION C)
Answer:
Option-D : They move freely in all directions.
Explanation:
The physical properties of gases are well explained by Kinetic Molecular Theory. The key postulates of this theory are;
1) Composition: Gases are made up of small particles called molecules. The size of these molecules is very small as compared to the distance between molecules, therefore the actual volume of molecules is taken negligible as compared to volume occupied by them.
2) Intermolecular Forces: All the gas molecules present in a container behaves independently because they have no force of interactions between them Hence, the attractive forces are taken negligible or too little.
3) Energies: Gas molecules have greater kinetic energy as compared to solids and liquids. Hence, The gas molecules move randomly. They collide with each other and with the walls of the container which causes pressure.
Conclusion:
Therefore, due to large spaces, no interactions, small sizes and high energies the gas particles move freely and there position is not stationary.
Answer:
At the end of second half life 12.5 g will left
Explanation:
Given data:
Total Mass = 50 g
Half lives = 2
Mass remain at the end = ?
Solution:
At time zero = 50 g
At 1st half life = 50 g /2 = 25 g
At second half life = 25 g/2 = 12.5 g
So at the end of second half life 12.5 g will left.
