Answer:
lesser the molar mass of the gas higher the no. of moles included in a certain mass sample. ie at STP more volume is required for the gas having less molar mass.
He has the smallest molar mass.
Therefore bag of He is the biggest.
The answer is Electron
Hope this helped!!! :)
The pH indicators to be used are Phenolphthalein, Red cabbage, Bromthymol blue and Congo red.
<h3>What are pH indicators?</h3>
Indicators are substances which change color as the pH of a medium changes.
The common indicators and their pH range is as follows:
- Phenolphthalein - pH range of 8.3 and 10.5
- Red cabbage - pH 2 to 10
- Bromthymol blue - 6.0 to 7.6
- Congo red - 3.0 to 5.2
Therefore, the indicators to be used are Phenolphthalein, Red cabbage, Bromthymol blue and Congo red.
Learn more about pH indicators at: brainly.com/question/13779537
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Answer:
Sc
:
4s
2
3d
1
Y
:
l
5s
2
4d
1
La
:
6s
2
5d
1
Ce
:
6s
2
4f
1
5d
1
Gd
:
6s
2
4f
7
5d
1
Lu
:
6s
2
4f
14
5d
1
Ac
:
7s
2
6d
1
Pa
:
7s
2
5f
2
6d
1
U
:
l
7s
2
5f
3
6d
1
Np
:
7s
2
5f
4
6d
1
Cm
:
7s
2
5f
7
6d
1
b. s2p3
he pnictogens:
N
l
:
2s
2
2p
3
P
l
:
3s
2
3p
3
As
:
4s
2
3d
10
4p
3
Sb
:
5s
2
4d
10
5p
3
Bi
:
6s
2
4f
14
5d
10
6p
3
Mc
:
7s
2
5f
14
6d
10
7p
3
c.The noble gases:
Ne
:
2s
2
2p
6
Ar
:
3s
2
3p
6
Kr
:
4s
2
3d
10
4p
6
Xe
:
5s
2
4d
10
5p
6
Rn
:
6s
2
4f
14
5d
10
6p
6
Og
:
7s
2
5f
14
6d
10
7p
6
Explanation: From the periodic tables we can drive elements with the electronic configuration
Sc
:
4s
2
3d
1
Y
:
l
5s
2
4d
1
La
:
6s
2
5d
1
Ce
:
6s
2
4f
1
5d
1
Gd
:
6s
2
4f
7
5d
1
Lu
:
6s
2
4f
14
5d
1
Ac
:
7s
2
6d
1
Pa
:
7s
2
5f
2
6d
1
U
:
l
7s
2
5f
3
6d
1
Np
:
7s
2
5f
4
6d
1
Cm
:
7s
2
5f
7
6d
1
b. s2p3
he pnictogens:
N
l
:
2s
2
2p
3
P
l
:
3s
2
3p
3
As
:
4s
2
3d
10
4p
3
Sb
:
5s
2
4d
10
5p
3
Bi
:
6s
2
4f
14
5d
10
6p
3
Mc
:
7s
2
5f
14
6d
10
7p
3
c.The noble gases:
Ne
:
2s
2
2p
6
Ar
:
3s
2
3p
6
Kr
:
4s
2
3d
10
4p
6
Xe
:
5s
2
4d
10
5p
6
Rn
:
6s
2
4f
14
5d
10
6p
6
Og
:
7s
2
5f
14
6d
10
7p
6
If I made no mistake in calculation, the given answer must be correct...(tried my best)
elements : carbon hydrogen oxygen Fluorine
composition [C] 24 3 16 57
M r 12 1 16 19
(divide C by Mr) 2 3 1 3
(Divide by smallest value) 2 3 1 3
(smallest value = 1...so all value remained constant)
Empirical formula : C2H3OF3
if molar mas = 100 g per mole, then
first step calculate Mr. of empirical formula: [= 100]
Them molecular formula = empirical formula
