Question

ks

Answer 1

this is suppose to be in chemistry

Answer 2

the molarity increases when the 2nd solution is added.
can you find the molarity of the 1st solution (in mg/ml)?

Answer 3

This is pre-calculus approach, not chemistry approach.

Answer 4

The formula for concentration is
\[\text{concentration} = \frac{\text{mass}}{\text{volume}}\]

Answer 5

So what you have to do first is write the concentration of the first beaker in terms of that.

Answer 6

Which is \[\frac{3\space\text{g}}{30\space \text{mL}}\] or simply
\[\frac{\text{g}}{10\space\text{mL}}\]

Answer 7

Next you have to write the concentration of the tank after adding n mL of the second solution to the first solution.

Answer 8

When you do that, you'll have the expression for a

Answer 9

The trick to doing that is to find the equivalent concentration of the given concentration of the second solution in terms of n.

In other words:

\[\frac{1 \space\text{mL}}{n \space \text{mL}} = \frac{8 \space\text{mg}}{x \space\text{mg}}\]

Answer 10

Solve that in terms of n. To make it easier, mL and mg cancel so isolate n for this proportion:

\[\frac{1}{n} = \frac{8}{x}\]

Answer 11

@jhalt, you there?

Answer 12

Solve for n. We're only on part (a) and we haven't finished that yet.

Answer 13

I'm still waiting on you to solve for n.

Answer 14

Sorry, why did I say solve for n? I mean, solve for x.

Answer 15

Sorry.

Answer 16

There you go

Answer 17

So if the concentration for the first beaker is \[\frac{3\space\text{g}}{30\space \text{mL}}\]

What will the concentration be if you add n mL to it?

Answer 18

After adding n mL of the second solution to the first solution, the new concentration will be

\[\frac{3 + 8n}{30 + n}\]

Answer 19

Finding part b is relatively easy now.

Answer 20

Part b simply says to let n = 10, and simplify the resulting expression.

Answer 21

Yes, correct.

Answer 22

Part C is somewhat of a trick question.

Answer 23

You have to convert 50 mg to grams

Answer 24

Since the new concentration is still in g/mL

Answer 25

So solve this:\[50 \space \text{mg} = x \space\text{g}\]

Answer 26

remember, m = 1/1000

Answer 27

and g cancels since you can just divide it out

Answer 28

Don't ask me to solve it for you. I'll just leave.

Answer 29

Just solve this:

\[\frac{50}{1000} = x\]

Answer 30

Well, it's already solved. Just simplify it

Answer 31

It simplifies further than that.

Answer 32

Good. So now solve this:

\[\frac{3 + 8n}{30 + n} = \frac{1}{20}\]

Answer 33

show me the work you did to get that.

Answer 34

@jhalt, please show me the work you did to get that. Just type your steps here.

Answer 35

Part c was asking something else

Answer 36

Hang on a bit

Answer 37

Okay, basically, I was stuck on this for a bit, but I figured out something about concentration formula

\[\text{concentration} = \frac{\text{mass of hydrogen peroxide}}{\text{volume of solution}}\]

Answer 38

So basically if you add anything to this concentration other than hydrogen peroxide, then the volume of solution will increase, but the mass of hydrogen peroxide will stay the same.

In other words, we will have to alter our original formula in order to solve this.

Answer 39

Since only the volume of solution will increase when we add water to the solution, V will represent the amount of water we add to the solution. So we have to solve this equation:

\[\frac{3}{30 + V} = \frac{1}{20}\]

Answer 40

So

60 = 30 + V

60 - 30 = V
30 = V

Answer 41

Therefore, we have to add 30 mL of water to the solution in order to obtain a 50 mg/mL solution.

Answer 42

For part D, all you need is a calculator.

Answer 43

and the expression we created for part a

Answer 44

\[\frac{3 + 8n}{30 + n}\]

Answer 45

Just keep increasing values of n by increments of 10 starting with 1 and just observe the behavior of the fraction as n increases. Then write what you observe about it.

Answer 46

@jhalt I'm never helping you with a math question again. You don't delete the question like that.

Answer 47

Because you're not supposed to delete your questions.

Answer 48

Doing so makes my replies look silly and also it prevents other students from being able to find this question should they seek the same explanation you did. I suggest that you put the question back. There were students who wanted to review this but couldn't because you deleted the question. They couldn't find this question and they certainly wouldn't be able to understand anything.

Answer 49

The question is, why did you delete the question in the first place?