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zimovet [89]
3 years ago
8

How does a scientist know where on the periodic table to put a newly discovered element?

Chemistry
1 answer:
deff fn [24]3 years ago
5 0

Explanation:

Periodic table has vertical columns and horizontals rows.

Periods:  The horizontal rows are named as periods in a periodic table.In a period elements with same numbers of principle quantum number ('n') or total number of shells (energy levels) are present. For example : Sodium comes in third period which means that sodium has three shells in total.

Groups: The vertical columns are named as group in a periodic table. In a group elements with same number electrons in their outermost shell are present. For example: In group one all the elements Li, Na, K etc have same number of valence electrons in their outermost shell.

So, in order to place a newly discovered element in a periodic table scientist must have the information regarding the number of valence electrons and total number of shells present in an atom of that element. One can easily look for valence number of electrons and total number of shells from the atomic number of that particular element.

By the knowledge of valence electrons and total number of shells the scientist will able to decide the group number and period number for newly discovered element.

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How many grams of NaOH are needed to make 250mL of a 0.1M NaOH solution?
Sloan [31]

Answer:

To make 250 ml of 0.1 M NaOH, you dissolve 1 gram NaOH in enough water to make a final volume of 250 mls.

Explanation:

Your welcome! :)

3 0
2 years ago
How many moles of each reactant are needed to produce 3.60 x 10^2g ch3oh
zysi [14]
Methanol is prepared by reacting Carbon monoxide and Hydrogen gas,

                                   CO  +  2 H₂    →    CH₃OH

Calculating Moles of CO:
                                       According to equation,

             32 g (1 mole) of CH₃OH is produced by  =  1 Mole of CO
So,
             3.60 × 10² g of CH₃OH is produced by  =  X Moles of CO

Solving for X,
                       X  =  (3.60 × 10² g × 1 Mole) ÷ 32 g

                       X  =  11.25 Moles of CO

Calculating Moles of H₂:
                                       According to equation,

             32 g (1 mole) of CH₃OH is produced by  =  2 Mole of H₂
So,
             3.60 × 10² g of CH₃OH is produced by  =  X Moles of H₂

Solving for X,
                       X  =  (3.60 × 10² g × 2 Mole) ÷ 32 g

                       X  =  22.5 Moles of H₂

Result:
            3.60 × 10² g of CH₃OH
is produced by reacting 11.25 Moles of CO and 22.5 Moles of H₂.
3 0
3 years ago
Loretta and her friends want to find out which of their pets can run the fastest. Which kind of scientific investigation should
Feliz [49]
A. and D. would be the best pick for this sort of experiment, but maybe (unlikely) B. because you could see how they could react in certain situations, how they react to danger but I suggest A.

Hope this helps you ☁︎☀︎☁︎
7 0
3 years ago
Read 2 more answers
A 52.0 mL portion of a 1.20 M solution is diluted to a total volume of 278 mL. A 139 mL portion of that solution is diluted by a
RoseWind [281]

Answer:

C_3=0.125M

Explanation:

Hello!

In this case, we can divide the problem in two steps:

1. Dilution to 278 mL: here, the initial concentration and volume are 1.20 M and 52.0 mL respectively, and a final volume of 278 mL, it means that the moles remain the same so we can write:

V_1C_1=V_2C_2

So we solve for C2:

C_2=\frac{C_1V_1}{V_2}=\frac{52.0mL*1.20M}{278mL}\\\\C_2=0.224M

2. Now, since 111 mL of water is added, we compute the final volume, V3:

V_3=139+111=250mL

So, the final concentration of the 139 mL portion is:

C_3=\frac{139 mL*0.224M}{250mL}\\\\C_3=0.125M

Best regards!

8 0
3 years ago
Which of the following is an irreversible physical change?
Yuliya22 [10]

Answer:

b) sharpening a pencil

Explanation:

If you melt lead, boil water, or dissolve sugar in water, you can return all of them back to their original state. If you sharpen a pencil, you can't reattach the shavings as they were originally.

3 0
3 years ago
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