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Sladkaya [172]
3 years ago
9

24. All elements found on the left side of the Periodic Table of

Chemistry
2 answers:
SashulF [63]3 years ago
5 0

Answer:

Metals are located on the left side of the periodic table and are generally shiny, malleable, ductile, and good conductors.

Explanation:

Ne4ueva [31]3 years ago
3 0

Answer:

have one delocalised electron

Explanation:

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A compound is 150.86 g carbon, 8.86 g hydrogen, and 20.10 g oxygen by
Ivahew [28]

Answer:

C20 H14 O2

Explanation:

Remark

This is a sample, which the question does not say and should. It is a fraction of 1 mole. So what you have to do is multiply the numbers given by x and equate it to 286.28

Equation

150,86* x + 8.86*x + 20.1*x  = 286.28

179.8x = 286.28

x = 286.26/179.8

x = 1.592

Now multiply the given numbers by 1.592

150.86 * 1.592 = 240.58

8.85 * 1.592 = 14.1

20.1 * 1.592 = 32

Rounding you get

240/12 = 20

14.1/1 = 14

32/16 = 2

C20 H14 O2

8 0
3 years ago
A binding protein binds to a ligand l with a kd of 400 nm. what is the concentration of ligand when y is (a) 0.25, (b) 0.6, (c)
lilavasa [31]

Hey there!:

The fractional saturation y is defined as :

y =  [ L ] / Kd + [ L ]

where :

[ L ] = concentration of binding ligand

Kd = 400 nm

3 0
3 years ago
I need to know the measurements of this to the appropriate amount of significant figures
igor_vitrenko [27]

Answer:

it can be 63 or 63.5

Explanation:

7 0
3 years ago
How many molecules are in 0.25 moles of CH4
Brums [2.3K]
The base gram of the compound is 16.042

To find the new amount of grams, you need to multiply the amount of moles by the the base grams

16.042*.25=4.01

So one fourth of the compound is 4.01 grams of Carbon tetrahydride.
6 0
3 years ago
If the energy difference between two electronic states is 214.68 kJ / mol , calculate the frequency of light emitted when an ele
atroni [7]

{\qquad\qquad\huge\underline{{\sf Answer}}}

Here we go ~

Energy difference btween the two electronic states can be expressed as :

{ \qquad \sf  \dashrightarrow \: \Delta E = h\nu}

[ h = planks constant,{\: \nu }= frequency ]

\qquad \sf  \dashrightarrow \:214.68 = 39.79 \times 10 {}^{ - 14}  \times  \nu

\qquad \sf  \dashrightarrow \: \nu =  \cfrac{214.68}{39.79 \times 10 {}^{ - 4} }

\qquad \sf  \dashrightarrow \: \nu =  \cfrac{214.68}{39.79 }  \times 10 {}^{14}

\qquad \sf  \dashrightarrow \: \nu  \approx  5.395 \times10 {}^{14}  \:\:hertz

5 0
2 years ago
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