Answer:
The mole ratio of the cation and the anion in a precipitate is a simple fraction. ( im sorry if this dosent help a lot.)
Explanation:
You should always do A. form a hypothesis before performing an experiment also the other options cannot happen until after an experiment.
Answer:
For these problems, we need to compare the theoretical yield that we'd get from performing stoichiometry to the actual yield stated in the problem. % yield is the actual yield/theoretical yield x 100%
Cu + 2 AgNO₠→ Cu(NOâ‚)â‚‚ + 2 Ag ==> each mole of copper yields two moles of silver
12.7-g Cu x ( 1 mol Cu /63.5-g Cu) x ( 2 mol Ag / 1 mol Cu) x (108-g Ag / 1 mol Ag) = 43.2-g Ag. This is the theoretical yield. Now, since we got 38.1-g Ag our % yield is:
38.1-g/43.2-g x 100% = 88.2%
Explanation:
Answer:
4 line graph
5 pictograph
10 vertical axis
2 bar graph
8 axis
3 pie chart
6 dependent variable
9 horizontal axis
1 graph
3 independent variable
Explanation:
Can I please have brainliest ;)
