Questions

[Mohit Pandey]

I have some questions to ask. It's not a quiz becoz I have doubts----

1.The value of e/m of an electron was measured by Millikan or J.J. Thomson?

2. CaC03 + 2HCl---CaCl2+ H2O+CO2

The mass 0f calcium chl0ride f0rmed WHEN 2.5g CaCO3 are dissolved in excess of HCl is __________. Explain.

[eric l]

Mohit, it is not that we do not want to answer some questions, rather we want to show you how to proceed for finding the answers yourself.

 

First, there is WIKIPEDIA (Wikipedia) a free multilingual encyclopedia. Though the answers it provides are not always sufficient or accurate, they are in 99 % of the cases. I often give a link to Wikipedia in my posts, for the simple reason that anyone who can post a question on Hypography, can search through Wikipedia (no subscription, no charges). And sometimes you have the possibility to check the information by selecting an other language. (the languages in which the same subject is treated appear in the column at the left).

 

Second, there is GOOGLE (and there are the other search engines like Yahoo and many others). If you go to Google (Google) there is a search window, in which you put - in this case electron + charge. This gives over 40 million hits (= items on the internet where both the words electron and charge are found). They are classified by relevance, so within the first 10 hits, you're bound to find one featuring Robert Milikan and his work on electron charge.

 

Thirdly : it is always a good idea to countercheck information from the internet - but that is always a good idea anyway.

 

Lots of success !

[Mercedes Benzene]

I quite agree. The answers to these questions can be easily deduced by means of an internet search.

You'll learn a lot more if you do things yourself. :pirate:

[Mohit Pandey]

OK. But the second one...:lol:

[eric l]

Where can you find the molecular weight of CaCO3 ? Either you have some table with atomic weights, and you add them up, or you check the internet. This will probably direct you to the Wikipedia article on calciumcarbonate. (No links this time - search for yourself).

Next do the same for CaCl2 or calciumchloride.

When you have both, you will see that 100 g of CaCO3 will react giving 111 g of CaCl2. From there it is basic arithmatic.

And if you have not yet learned how to calculate the molecular weight of a substance of which you have the formula, you are playing out of your league, and have to be patient for some months.

[Mohit Pandey]

One more questi0n. :)This time I have g0ne t0 G00gle. But it didn't pr0vide the answer.

Which 0f the f0ll0wing electr0ns transiti0n in hydrogen require the largest amount of energy?

1.[math]n_L =1 [/math] to [math] n_H=2 [/math]

2..[math]n_L =2 [/math] to [math] n_H=3 [/math]

3.[math]n_L =1 [/math] to [math] n_H=infinity [/math]

4..[math]n_L =3 [/math] to [math] n_H=5 [/math]

 

Please tell the concept involved in simple terms .

[eric l]

I do not want to be patronizing or worse, but if you Google

hydrogen + "electron transition"

you get a number of links.

One of the first is this :

Hydrogen energies and spectrum

I do not blame you for not understanding the content, it is the kind of stuff I also need to read over and over to digest. But I'm affraid this is simply out of your league. From what I learned about you, you are in high school (or equivalent) starting to learn chemistry, and quite enthousiastic about that. And I applaud that. But this energy levels of electron transitions are more university stuff.

[Mohit Pandey]

I do not blame you for not understanding the content, it is the kind of stuff I also need to read over and over to digest. But I'm affraid this is simply out of your league. From what I learned about you, you are in high school (or equivalent) starting to learn chemistry, and quite enthousiastic about that. And I applaud that. But this energy levels of electron transitions are more university stuff.

 

But you have the quality to explain in simple terms as you did in thread 'Greater no.':thumbs_up. I hope you can explain this question also.