You have to put your attention to the unit of concentration. It is expressed in terms of molarity, which is represented in M. It is the number of moles solute per liter solution. So, you simply have to multiply the molarity with the volume in liters.
Volume = 275 mL * 1 L/1000 mL = 0.275 L
<em>Moles Ba(OH)₂ = (0.200 M)(0.275 L) = 0.055 mol</em>
B is the.answer for this problem
Given:
No of atoms present= 8.022 x 10^23 atoms
Now we know that 1 mole= 6.022 x 10^23 atoms
Hence number of moles present in 8.022 x 10^23 atoms is calculated as below.
Number of moles
= 8.022 x 10^23/6.022x 10^23
=1.3 moles.
Hence we have 1.3 moles present.
Answer: The mass of given amount of copper (II) cyanide is 462.4 g
Explanation:
To calculate the number of moles, we use the equation:
We are given:
Moles of copper (II) cyanide = 4 moles
Molar mass of copper (II) cyanide = 115.6 g/mol
Putting values in above equation, we get:
Hence, the mass of given amount of copper (II) cyanide is 462.4 g