Somewhat false
observations can be made of a model of the statue of liberty, say, or in real line
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:
Bacteria
Explanation:
hope this helps please gimme brainliest
Answer:
A 12 oz Coca Cola contains 39g of sugar or C6H12O6.
To calculate for the molarity of sugar in the soda, convert 39 grams of sugar to moles sugar:
39g/ 180.16 g/mol = 0.216 mol sugar
then, convert 12 oz to L:
12oz / (1oz/0.02957L) = 0.35484 L
therefore the concentration of sugar in the soda is:
M = mol sugar / L sol'n
= 0.216 mol sugar / 0.35484 L
= 0.609 M
Explanation:
In this item, we are simply to find the ions that may bond and are able to form a formula unit. We are also instructed to give out their name. There are numerous possible combinations of ions to form a compound. Some answers are given in the list below.
1. Na⁺ , Cl⁻ , NaCl ---> sodium chloride (this is most commonly known as table salt)
2. C⁴⁺ , O²⁻ , CO₂ ---> carbon dioxide
3. Al³+ , Cl⁻ , AlCl₃ ----> aluminum chloride
4. Ca²⁺ , Cl⁻ , CaCl₂ ---> calcium chloride
5. Li⁺ , Br⁻ , LiBr ---> lithium bromide
6. Mg³⁺ , O²⁻ , Mg₂O₃ ----> magnesium oxide
7. K⁺ , I⁻ , KI ---> potassium iodide
8. H⁺ , Cl⁻ , HCl --> hydrogen chloride
9. H⁺ , Br⁻ , HBr ----> hydrogen bromide
10. Na⁺ , Br⁻ , NaBr ---> sodium bromide
