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qwelly [4]
3 years ago
7

27. What element has this electron configuration: 1s2 2s22p6 352 3p6 452

Chemistry
1 answer:
stira [4]3 years ago
4 0

Answer:

Selenium

Explanation:

One way to figure this out is to count the electrons and add them up.

2+2+6+2+6+2+10+4 = 34 electrons

A neutral atom has the same number of electrons and protons. Protons tell us the atomic number of an element. Element number 34 is Selenium.

Another way to figure it out is to look at just the highest "s" or "p" level of the electron configuration, and then see which element corresponds to that. This element has a 4s and a 4p, so I will look at 4p. The configuration is 4p4, so I will look at the 4th row of the periodic table (4p4) in the "p" block (4p4) and the 4th element in the "p" block (4p4). That's selenium :)

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Knowing the ratio between atoms we can write an empirical formula: 

<span>C4H6O </span>

<span>we compute the molar mass of this single formula: </span>
<span>4x12 + 6 x 1 + 16 x1 = 70 g / mol </span>

<span>Now, as we know the actual molar mas being 280 g/mol, we divide this number by 70 and we get the ratio between empirical formula and molecular actual formula: </span>

<span>280 / 70 = 4 </span>

<span>This means that actual molecular formula is: </span>

<span>(C4H6O)4 or </span>

<span>C16H24O4 </span>
3 0
3 years ago
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The electron configuration of bromine is 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5 classify electrons in each.
bulgar [2K]

Bromine has the following electron configuration: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5. categorize the electrons in each. Answer for video: The video player is loaded.

On the periodic chart, row 5, column 7, is where you can find a chemical element that was identified in 1811. It has a proton count of 53 and an atomic mass of 126.9. Iodine's atom, then, contains 53 electrons in the following configuration: 1s2, 2s2, 2p6, 3s2, 3d10, 4p6, 5s2, 4d10, 5p5 (Kr 4d10 5s2 5p5). Cu Z = 29 has an electrical arrangement of 1s2 2s2 2p6 3s2 3p6 3d10 4s1. Copper (Co) has the following electron configuration: 1s2 2s2 2p6 3s3 3p6 4s2 3d7. If a chemist were to refer to Copper by its subshell, they would abbreviate this notation to "3d7."

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7 0
1 year ago
If 22.5L of nitrogen at 748 mm Hg are compressed to 725 mm Hg at constant temperature. What is the new volume?2.A gas with a vol
kykrilka [37]

Answer:

a)23.2 L

b)68.3kPa

c)7.5 atm

d)60.5L

e)1.67 atm

Explanation:

From Boyle's law:

P1V1=P2V2

P1= 748mmHg

P2=725mmHg

V1= 22.5L

V2??

V2= P1V1/P2= 748×22.5/725= 23.2 L

b)

V1=4.0L

P1= 205×10^3Pa

V2= 12.0L

P2=???

P2= P1V1/V2= 205×10^3×4/12

P2= 68.3×10^3 Pa or 68.3kPa

c)

P1= 1 atm

V1= 196.0L

P2= ??

V2= 26.0L

P2= P1V1/V2=1×196.0/26.0

P2= 7.5 atm

d)

V1= 40.0L

P1= 12.7×10^3Pa

V2=???

P2= 8.4×103Pa

V2= P1V1/P2= 12.7×10^3×40.0/8.4×103

V2=60.5L

e)

V1= 100mL

P1= 1atm

V2= 60mL

P2=???

P2= P1V1/V2= 1×100/60

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4 years ago
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jekas [21]

Answer:

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4 0
3 years ago
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larisa86 [58]

Answer:

+15.8°

Explanation:

The formula for the observed rotation (α) of an optically active sample is

α = [α]<em>lc </em>

where

<em>l</em> = the cell path length in decimetres

<em>c</em> = the concentration in units of g/100 mL

[α] = the specific rotation in degrees  

1. Convert the concentration to units of g/100 mL

c = \frac{\text{0.119 g}}{\text{10.0 mL}} \times \frac{10 }{10 } =\frac{\text{1.19 g}}{\text{100 mL}}

2. Calculate the observed rotation

\alpha = (+13.3 ^{\circ}) \times 1.19 \times 1 = +15.8^{\circ}

7 0
4 years ago
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