Radiant energy or electrical energy
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.
Answer:
F - O - S - Mg - Ba
Explanation:
as you move left to right on the periodic table the number of electrons increase.
The answer to your question is A. Wrought iron because amongst its other properties, wrought iron becomes soft at red heat, and can be easily be forged and forge welded.
<u>Answer:</u> The moles of hydroxide ions present in the sample is 0.0008 moles
<u>Explanation:</u>
To calculate the concentration of acid, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is HCl.
are the n-factor, molarity and volume of base which is 
We are given:
Putting values in above equation, we get:
To calculate the moles of hydroxide ions, we use the equation used to calculate the molarity of solution:
Molarity of solution = 0.011 M
Volume of solution = 36.0 mL
Putting values in above equation, we get:
1 mole of calcium hydroxide produces 1 mole of calcium ions and 2 moles of hydroxide ions.
Moles of hydroxide ions = (0.0004 × 2) = 0.0008 moles
Hence, the moles of hydroxide ions present in the sample is 0.0008 moles
