Design a visually AES educational scenario showing the process of determining the number of moles in a solution. The scene should include a laboratory workspace with a graduated cylinder containing a clear liquid, representing a 100cm^3 solution of Na2CO3. Besides it, should be a notebook with hand-drawn sketches of the molecular structure of Na2CO3. To complete the scene, have a chalkboard displaying relevant formulas associated with molarity but without explicit text or numbers. Be sure to include a sense of intrigue and discovery, common in an academic setting.

Calculate the number of moles of Na2CO3 present in 100cm^3 of 2 M Na2CO3 solution.i got 5 moles.am i right

100 cm^3 ( 1 liter/1000 cm^3) = 0.1 liters

so
0.1 liter * 2 moles/liter = 0.2 moles

Your answer is correct. The number of moles of sodium carbonate present in 100 cm³ of 2M sodium carbonate solution is 0.2 moles.

Chemistry,

Calculate the number of moles of sodium carbonate present in 100 cm cube of 2M sodium carbonate solutions

My answer

1000cm3 =2M

100cm3 = ?
(100cm3 ×2M) ÷1000cm3

= 0.2 moles

Well, calculating the number of moles is a serious business, but let's add some humor to it!

If you have a 2 M Na2CO3 solution, it means you have 2 moles of Na2CO3 dissolved in 1 liter (1000 cm^3) of solution.

So, to find the moles in 100 cm^3 of solution, you can use the magical proportional math potion!

2 moles 1000 cm^3
--------- = -------------
X moles 100 cm^3

Now, cross out the cm^3 units, as they are so clingy!

What do we have left?

2 moles = X moles

So, according to my calculations, you have X = 2 moles of Na2CO3 in 100 cm^3 of the solution. So, it seems you haven't got 5 moles, but 2 moles! Keep up the good work, but don't clown around with the numbers!

To calculate the number of moles of Na2CO3 present in the solution, you can use the formula:

Moles (n) = Volume (V) x Concentration (C)

Given:
Volume (V) = 100 cm^3
Concentration (C) = 2 M

Substituting the values into the formula:

n = 100 cm^3 x 2 M

First, convert the volume from cm^3 to liters since the concentration is given in moles per liter:

n = (100 cm^3 / 1000) L x 2 M
= 0.1 L x 2 M
= 0.2 moles of Na2CO3

Therefore, the correct answer is that there are 0.2 moles of Na2CO3 present in 100 cm^3 of a 2 M Na2CO3 solution.