Answer:
3 e⁻ transfer has occurred.
Explanation
This is a redox reaction.
- Oxidation (loss of electrons or increase in the oxidation state of entity)
- Reduction (gain of electrons or decrease in the oxidation state of the entity)
- An element undergoes oxidation or reduction in order to achieve a stable configuration. It can be an octet or duplet configuration. An octet configuration is that of outer shell configuration of noble gas.
- [Ne]= (1s²) (2s² 2p⁶)
A combination of both the reactions( Half-reactions) leads to a redox reaction.
Let us look at initial configurations of Al and Cl
[Al]= 1s² 2s² 2p⁶ 3s² 3p¹
[Cl]= 1s² 2s² 2p⁶ 3s² 3p⁵
Hence, Al can lose 3 electrons to achieve octet config.
and, Cl can gain 1e to achieve nearest noble gas config. [Ar]
This reaction can be rewritten, by clearly mentioning the oxidation states of all the entities involved.
Al⁰ + Cl⁰ → (Al⁺³)(Cl⁻)₃
Here, Aluminum is undergoing an oxidation(i.e loss of electrons) from: 0→(+3)
Chlorine undergoes a reduction half reaction (i.e gain of electrons) from: 0→(-1). There are 3 such chlorine atoms, hence 3 e⁻ transfer has occurred.
False. Electromagnetic and Gravitational forces can overcome .
Answer:
The half-life of the radioactive isotope is 346 years.
Explanation:
The decay rate of the isotope is modelled after the following first-order linear ordinary differential equation:
Where:
- Current isotope mass, measured in kilograms.
- Time, measured in years.
- Time constant, measured in years.
The solution of this differential equation is:
Where 
is the initial mass of the isotope. It is known that radioactive isotope decays at a yearly rate of 0.2 % annually, then, the following relationship is obtained:
The time constant associated to the decay is:
Finally, the half-life of the isotope as a function of time constant is given by the expression described below:
If , the half-life of the isotope is:
The half-life of the radioactive isotope is 346 years.
Glucose falls under aldoses group of monosaccharides which are one of the carbohydrates.
<u>Explanation:</u>
Generally, a molecule formed by combination of carbon, hydrogen and oxygen are termed as carbohydrates. The smallest of the saccharides are mono and disaccharides. The saccharides can contain an aldehyde or a ketone group in their structural formula.
The molecular formula of glucose is C6H12O6. It falls under monosaccharide as the molecular formula of monosaccharides are (CH2nO)n where the n will be 3,5 or 6. So the molecular formula of glucose exactly matches with the molecular formula of monosaccharide family.
Also, it is known that glucose contains an aldehyde so it can be termed as aldose for being a carbohydrate in monosaccharide family containing aldehyde group.
Answer:
159 g OF LiNO2 WILL BE USED TO MAKE 0.25 L OF 0.75 M SOLUTION.
Explanation:
How many grams of LiNO2 are required to make 250 mL of 0.75 M?
First calculate the molarity in mol per dm3
0.75 M of LiNO2 reacts in 250 mL = 250 /1000 L volume
0.75 M = 0.25 L
In 1 L, the molarity of LiNO2 will be:
= (0.75 * 1/ 0.25) M
= 3 mol/dm3 of LiNO2
Next is to calculate the molarity in g/dm3:
Molarity in mol/dm3 = molarity in g/dm3 / RMM.
RMM of LiNO2
(Li = 7 , N =14 , 0 = 16)
RMM = ( 7 + 14 + 16 * 2) = 53 g/mol
Molarity in mol/dm3 = Molarity in g/dm3 / RMM
Molarity in g/dm3 = Molarity in mol/dm3 * RMM
Molarity in g/dm3 = 3 * 53
Molarity in g/dm3 = 159 g/dm3.
So therefore, to make 250 mL of 0.75 M of LiNO2, we use 159 grams of LiO2.
