Answer:
Explanation:
Hello,
In this case, for the sample of the given compound, we can compute the moles of each atom (carbon, hydrogen and oxygen) that is present in the sample as shown below:
- Moles of carbon are contained in the 9.582 grams of carbon dioxide:
- Moles of hydrogen are contained in the 3.922 grams of water:
- Mass of oxygen is computed by subtracting both the mass of carbon and hydrogen in carbon dioxide and water respectively from the initial sample:
Finally, we compute the percent by mass of oxygen:
Regards.
Answer:
the solvent is water and the solute is salt
Explanation:
A solute is the component in a solution in the lesser amount. In a NaCl solution, the salt is the solute. A solution may contain more than one solute.
Answer:
Mass = 112.54 g
Explanation:
Given data:
Mass of copper = 18 g
How much copper(II) nitrate formed = ?
Solution:
Cu + 2AgNO₃ → Cu(NO₃)₂ + 2Ag
Number of moles of copper:
Number of moles = mass/ molar mass
Number of moles = 18 g/ 29 g/mol
Number of moles = 0.6 mol
Now we will compare the moles of Cu with Cu(NO₃)₂ .
Cu : Cu(NO₃)₂
1 : 1
0.6 : 0.6
Mass of Cu(NO₃)₂ :
Mass = number of moles × molar mass
Mass = 0.6 mol × 187.56 g/mol
Mass = 112.54 g
A) can enter from the surroundings, but cannot escape to the surroundings
Answer:
Oxidation half reaction is written as follows when using using reduction potential chart
example when using copper it is written as follows
CU2+ +2e- --> c(s) +0.34v
oxidasation is the loos of electron hence copper oxidation potential is as follows
cu (s) --> CU2+ +2e -0.34v
Explanation:
