Want this question answered?
Well, first you start with a problem . . .
You don't solve stoichiometry. The questions and answers that arise in stoichiometry are merely manipulations of permanent relationships between things (e.g. there are approximately 70.9 grams in one mole of chlorine gas). The conversions needed to report an answer of a stoichiometric problem are the part that take work to overcome mentally. One has to evaluate the units that a value starts with and the units the final answer requires and think about what conversions are needed in between.
Depends, proportional factor could be a method.
Moles
This is a complicated titration problem. The volumes are necessary to solve the problem, as well as a Ka or Kb value.
Well, first you start with a problem . . .
Understand the problem by reading it carefully and clarifying any uncertainties. Plan a solution by breaking down the problem into smaller tasks or steps. Execute the plan by following the steps you've identified. Evaluate the solution to ensure it addresses the problem effectively and make adjustments if needed.
You don't solve stoichiometry. The questions and answers that arise in stoichiometry are merely manipulations of permanent relationships between things (e.g. there are approximately 70.9 grams in one mole of chlorine gas). The conversions needed to report an answer of a stoichiometric problem are the part that take work to overcome mentally. One has to evaluate the units that a value starts with and the units the final answer requires and think about what conversions are needed in between.
Depends, proportional factor could be a method.
it is a prediction at what you think will be the outcome of the experiment...
Moles
There's no way to solve this problem....
No. We solve problems with algorithms, not with syntax.
Describe varios steps necessary to solve a problem
Keep working the problem until you solve it. Get some help if necessary, but not during a test!
4t
This is a complicated titration problem. The volumes are necessary to solve the problem, as well as a Ka or Kb value.