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Elan Coil [88]
3 years ago
9

Which is the correct Lewis structure of C2H4

Chemistry
1 answer:
Maslowich3 years ago
3 0
Lewis structure for C2H4

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Calculate the extinction coefficient where the concentration is in mg/ml and the path length is 1 cm. What dilutions of the stoc
Airida [17]

Complete Question

The complete question is show on the first uploaded image

Answer:

This is shown on the second,third , fourth and fifth image

Explanation:

This is shown on the second,third , fourth and fifth image

4 0
3 years ago
Mercury has a density of 13.6 grams/mL, what is this density in cg/L?
Firdavs [7]
It would be 4.6cgL not sure tho because I didint do that good in this
4 0
2 years ago
Which of the following combination of elements would result in covalent compound? * W X Y Z Vand X Wand Z Y and Z Wand y​
TEA [102]

Answer:

C. Y & Z

Explanation:

V, W are imaginary metals here because their valence electrons are typically less than 4. X, Y, Z are non-metals and have higher valence electrons. Here, if V or W bind with X, Y, or Z we make ionic bond (because metal + non metal = ionic). But, if X binds with Y or Z or any combinations of any two of the three non-metals results in covalent bond (non metal + non metal = covalent).

Thus, Y and Z make covalent.

4 0
2 years ago
An atom of a particular element is traveling at 1.00% of the speed of light. The de Broglie wavelength is found to be 3.31 × 10-
ExtremeBDS [4]

Answer:

The given atom is of Ca.

Explanation:

Given data:

Speed of atom = 1% of speed of light

De-broglie wavelength = 3.31×10⁻³ pm (3.31×10⁻³ / 10¹² = 3.31×10⁻¹⁵ m)

What is element = ?

Solution:

Formula:

m = h/λv

m = mass of particle

h  = planks constant

v = speed of particle

λ = wavelength

Now we will put the values in formula.

m = h/λv

m = 6.63×10⁻³⁴kg. m².s⁻¹/3.31×10⁻¹⁵ m ×( 1/100)×3×10⁸ m/s

m = 6.63×10⁻³⁴kg. m².s⁻¹/ 0.099×10⁻⁷m²/s

m = 66.97×10⁻²⁷ Kg/atom

or

6.69×10⁻²⁶ Kg/atom

Now here we will use the Avogadro number.

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

18 g of water = 1 mole = 6.022 × 10²³ molecules of water

Now in given problem,

6.69×10⁻²⁶ Kg/atom × 6.022 × 10²³ atoms/ mol × 1000 g/ 1kg

40.3×10⁻³×10³g/mol

40.3  g/mol

So the given atom is of Ca.

8 0
2 years ago
Independent variable ​
katrin [286]

Answer:

Independent variable

aluminium ball

6 0
3 years ago
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