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andre [41]
3 years ago
9

Lithium (Li) and oxygen (O) are both in period 2. They both have _____.

Chemistry
1 answer:
AlladinOne [14]3 years ago
6 0

They both have two electron shells

<h3>Further explanation</h3>

The period 2 element lies in the second row of the periodic system.

Consists of the elements: lithium, beryllium, boron, carbon, nitrogen, oxygen, fluorine, and neon

  • Lithium (Li)

atomic number : 3

electron configuration : [He] 2s¹

atomic number = number of proton=number of electron(in neutral atom)

So Li have 3 protons and 3 electrons

Because it fills the 2s orbital it has 2 shells

  • Oxygen (O)

atomic number : 8

electron configuration :  [He] 2s²2p⁴

So O have 8 protons and 8 electrons

Because it fills the 2s and 2p orbital it has 2 shells

So Lithium (Li) and Oxygen (O) are both have two electron shells

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7. Which of the following would lower the pitch of a string instrument's sound?
vfiekz [6]

The correct option is C.

The pitch of a string refers to the quality of sound that is produced by the string when it vibrates. There are three basic factors that affect the quality of pitch of a string, these are: the tension, the thickness of the string and the length of the string. The higher the thickness of the string, the lower the pitch of the string, thus, increasing the thickness of the string will lower the pitch.

3 0
3 years ago
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A compound with the empirical formula ch2 has a molar mass of 70 g/mol. what is the molecular formula for this compound?
Rainbow [258]

The empirical formula, <span>C<span>H2</span></span>, has a relative molecular mass of

<span>1×<span>(12.01)</span>+2×<span>(1.01)</span>=14.04</span>

This means that the empirical formula must be multiplied by a factor to bring up its molecular weight to 70. This factor can be calculated as the ratio of the relative masses of the molecular and empirical formulas

<span><span>7014.04</span>=4.98≈5</span>

Remember that subscripts in molecular formulas must be in whole numbers, hence the rounding-off. Finally, the molecular formula is

<span><span>C<span>1×5</span></span><span>H<span>2×5</span></span>=<span>C5</span><span>H<span>10</span></span></span>

5 0
4 years ago
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The raw water supply for a community contains 18 mg/L total particulate matter. It is to be treated by addition of 60 mg alum (A
s344n2d4d5 [400]

Solution :

Given :

The steady state flow = 8000 $ m^3 /d $

                                    $= 80 \times 10^5 \ I/d $

The concentration of the particulate matter = 18 mg/L

Therefore, the total quantity of a particulate matter in fluid $= 80 \times 10^5 \ I/d \times 18 \ mg/L $

$= 144 \times 10^6 \ mg/g$

$= 144 \ kg/d $

If 60 mg of alum $ [Al_2(SO_4)_3.14 H_2O] $ required for one litre of the water treatment.

So Alum required for  $ 80 \times 10^5 \ I/d $

$= 80 \times 15^5 \ I/d  \times 60 \ mg \ alum /L$

$= 480 \times 10^6 \ mg/d $

or 480 kg/d

Therefore the alum required is 480 kg/d

1 mg of the alum gives 0.234 mg alum precipitation, so 60 mg of alum will give $ = 60 \times 0.234 \text{ of alum ppt. per litre} $

      $= 14.04 $ mg of alum ppt. per litre

480 kg of alum will give = 480 x 0.234 kg/d

                                        = 112.32 kg/d ppt of alum

Daily total solid load is  $= 144 \ kg/d + 112.32 \ kg/d$

                                       = 256.32 kg/d

So, the total concentration of the suspended solid after alum addition $= 18 \ mg/L + 60 \times 0.234 $

= 32.04 mg/L

Therefore total alum requirement = 480 kg/d

b). Initial pH = 7.4

 The dissociation reaction of aluminium hydroxide as follows :

$Al(OH)_3 \rightleftharpoons Al^{3+} + 3OH^{-} $

After addition, the aluminium hydroxide pH of water will increase due to increase in $ OH^- $ ions.

Therefore, the pH of water will be acceptable range after the addition of aluminium hydroxide.

c). The reaction of $CO_2$ and water as follows :

$CO_2 (g) + H_2O (l) \rightarrow H_2CO_3$

For the atmospheric pressure :

$p_{CO_2} = 3.5 \times 10^{-4} \ atm $

And the pH is reduced into the range of 5.9 to 6.4

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3 years ago
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irina1246 [14]

Answer:

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The element in Period 4 and Group 14 of the Periodic Table would be classified as a
Paul [167]

Answer:

Chalcogen

Explanation:

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