Answer:
<h2>117.94 moles</h2>
Explanation:
To find the number of moles in a substance given it's number of entities we use the formula
where n is the number of moles
N is the number of entities
L is the Avogadro's constant which is
6.02 × 10²³ entities
From the question we have
We have the final answer as
<h3>117.94 moles</h3>
Hope this helps you
There are many properties to substances.
I'll list some examples below:
- Mass
- Volume
- Density
- Conductivity
- Malleability
- Boiling point
- Melting point
- Heat capacity
Hope this helps! :3
Answer:
Oxygen is a simple molecular structure, where individual oxygen atoms are bonded to each other by strong covalent bonds. Hence, a low amount of energy is required to overcome these weak forces and oxygen has a low boiling point. Therefore, at room temperature, oxygen is a gas. Oxygen difluoride is a colorless gas, condensable to a pale yellow liquid, with a slightly irritating odor. It is the most stable of the compounds of fluorine and oxygen, which include O,F,, O,F, and 0,F2 but nevertheless it is a strong oxidizing and fluorinating agent. Oxygen Difluoride is a colorless gas or a yellowish-brown liquid with a foul odor. Just to finally link Joseph's answer to the question, oxygen difluoride will thus change from liquid to solid state when chilled from -220°c to -230°c. The boiling point of oxygen is -182.96 degrees Celsius (under 1 standard atmosphere). This means at temperatures below that point, oxygen is a solid or a liquid, and at temperatures above that point, oxygen is a gas. So at -183 degrees Celsius, oxygen is a liquid.
Explanation:
Explanation:
Molarity is defined as number of moles per liter of solution.
Mathematically, molarity = 
It is given that molarity is 0.0800 M and volume is 50.00 mL or 0.05 L.
molarity = 
0.0800 M = 
no. of moles = 1.6 mol
Therefore, molar mass of cupric sulfate pentahydrate is 249.68 g/mol. So, calculate the mass as follows.
No. of moles = 
mass in grams = 
= 
= 399.488 g
Thus, we can conclude that 399.488 g of cupric sulfate pentahydrate are needed to prepare 50.00 mL of 0.0800M CuSO4× 5H2O.
