The Beer-Lambert law states that A = E*c*l where A is absorbance, E is the molar absorbance coeffecient, c is concentration and l is path length. Therefore the absorbance is directly proportional to concentration, and by increasing the concentration by a factor of 3, absorbance will increase by a factor of 3 giving A = 1.584
While I am not the brainliest I can certainly answer.
This was a chemical change because the chemical components were changed, a big giveaway to this was the fizzing, however the temperature rising was also another giveaway.
Explanation:
<u>Physical properties of ZBr₂</u>
The compound is an ionic substance. Therefore it will have properties of ionic compounds. Some of these properties are:
- it is a hard solid usually with high melting points or a liquid with high boiling points.
- Soluble in water and insoluble in non-polar solvents.
- It can conduct electricity in aqueous solutions or in molten form.
- it will undergo a fast reaction.
<u>Z is a metal</u>
To form ionic compound, a metal will combine with a non-metal. Bromine is a non-metal and it is expected that Z will be a metal. This is because ionic bonds involves transfer of electron from one specie to the other. Metals are usually the donor and non-metals are the receptor. This is how ionic bond forms. The electrostatic attraction resulting from the ions produced the ionic bond.
<u>Formula of the oxide</u>
ZO
Z 0
+2 -2
It is obvious that Z has 2 valence electrons. It will lose the two valence electrons to attain stability.
Oxygen requires 2 electrons to resemble Neon. This combination will give a compound ZO.
Learn more:
ionic compounds brainly.com/question/6071838
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Answer:
4
Explanation:
Relationship between wavenumber and Rydberg constant (R) is as follows:
Here, Z is atomic number.
R=109677 cm^-1
Wavenumber is related with wavelength as follows:
wavenumber = 1/wavelength
wavelength = 253.4 nm
Z fro Be = 4
Therefore, the principal quantum number corresponding to the given emission is 4.
Limiting reactant : O₂
Mass of N₂O₄ produced = 95.83 g
<h3>Further explanation</h3>
Given
50g nitrous oxide
50g oxygen
Reaction
2N20 + 302 - 2N204
Required
Limiting reactant
mass of N204 produced
Solution
mol N₂O :
mol O₂ :
2N₂O+3O₂⇒ 2N₂O₄
ICE method
1.136 1.5625
1.0416 1.5625 1.0416
0.0944 0 1.0416
Limiting reactant : Oxygen-O₂
Mass N₂O₄(MW=92 g/mol) :
