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DedPeter [7]
3 years ago
6

What is the purpose of the stairstep pattern on the periodic table of the elements?

Chemistry
1 answer:
Mamont248 [21]3 years ago
7 0

Answer: The bold staircase in the periodic table allows us to classify which elements are metalloids.

Explanation: Additionally, it acts like a "divider" that allows us to properly distinguish the metals from the non-metals in the periodic table.

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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
Law of Conservation of Mass in terms of atoms
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4 0
3 years ago
Name the following alkyne: CH3CH₂C = CCH₂CH₂CH3
liubo4ka [24]

The given alkyne is Option A 3-heptyne

<h3>What is an Alkyne ?</h3>

The hydrocarbon having at least one C-C triple bond is called an Alkyne.

It has the general formula of   \rm C_n H_{2n+2} .

In the question it is being mentioned that it is an alkyne so there will be a triple bond and not a double bond.

It has been asked in the question that

CH3CH₂C ≡  CCH₂CH₂CH3 is which alkyne from the given option.

The counting of the Carbon chain is done from the left side and the Triple bond is at the 3rd Carbon , so 3-heptyne .

To know more about Alkyne

brainly.com/question/23508203

#SPJ1

<h3 />

6 0
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What evidence would support Maria's claim that the fossilized remains are that of a bird, class Aves, instead of a turtle, class
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Your answer would be C.

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