Answer:
32.1 × 10²³ formula units of CuCl₂
Explanation:
Given data:
Number of moles of CuCl₂ = 5.33 mol
Number of formula units of CuCl₂ = ?
Solution:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
one mole of any substance contain 6.022 × 10²³ formula units thus,
5.33 moles of CuCl₂ = 5.33 ×6.022 × 10²³ formula units
32.1 × 10²³ formula units of CuCl₂
Answer:
<em>yh thats true lol, ty for that very interesting fact</em>
Answer: The molarity of KBr in the final solution is 1.42M
Explanation:
We can calculate the molarity of the KBr in the final solution by dividing the total number of moles of KBr in the solution by the final volume of the solution.
We will first calculate the number of moles of KBr in the individual sample before mixing together
In the first sample:
Volume (V) = 35.0 mL
Concentration (C) = 1.00M
Number of moles (n) = C × V
n = (35.0mL × 1.00M)
n= 35.0mmol
For the second sample
V = 60.0 mL
C = 0.600 M
n = (60.0 mL × 0.600 M)
n = 36.0mmol
Therefore, we have (35.0 + 36.0)mmol in the final solution
Number of moles of KBr in final solution (n) = 71.0mmol
Now, to get the molarity of the final solution , we will divide the total number of moles of KBr in the solution by the final volume of the solution after evaporation.
Therefore,
Final volume of solution (V) = 50mL
Number of moles of KBr in final solution (n) = 71.0mmol
From
C = n / V
C= 71.0mmol/50mL
C = 1.42M
Therefore, the molarity of KBr in the final solution is 1.42M
Answer:
Percent error = 12.5%
Explanation:
In a measurement you can find percent error following the formula:
Percent error = |Measured value - Accepted Value| / Acepted value * 100
Based on the data of the problem, accepted value is 22.4L and the measured Value (Value of Sara) was 19.6L.
Replacing:
Percent error = |Measured value - Accepted Value| / Acepted value * 100
Percent error = |19.6L - 22.4L| / 22.4L * 100
Percent error = |-2.8L| / 22.4L * 100
Percent error = 2.8L / 22.4L * 100
Percent error = 12.5%
Answer:
c
O2: 7 mol
CO2: 4 mil
