Answer:
1.45 x 10⁻² g CO₂
Explanation:
To find the mass of carbon dioxide, you need to (1) convert grams CH₄ to moles CH₄ (via molar mass), then (2) convert moles CH₄ to moles CO₂ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles CO₂ to grams CO₂ (via molar mass). The final answer should have 3 sig figs to reflect the given value (5.30 x 10⁻³ g).
Molar Mass (CH₄): 12.011 g/mol + 4(1.008 g/mol)
Molar Mass (CH₄): 16.043 g/mol
Combustion of Methane:
1 CH₄ + 2 O₂ ---> 2 H₂O + 1 CO₂
Molar Mass (CO₂): 12.011 g/mol + 2(15.998 g/mol)
Molar Mass (CO₂): 44.007 g/mol
5.30 x 10⁻³ g CH₄ 1 mole 1 mole CO₂ 44.007 g
--------------------------- x ---------------- x --------------------- x ----------------- =
16.043 g 1 mole CH₄ 1 mole
= 0.0145 g CO₂
= 1.45 x 10⁻² g CO₂
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.
It is a liquid because when you have a liquid, there is no definite shape. Therefore, this would be the answer because it takes the shape of its container.
Final answer: a. Liquid
Answer:
The answer is E. All of the statements describe the anomeric carbon.
Explanation:
When a sugar switches from its open form to its ring form, the carbon from the carbonyl (aldehyde if it is an aldose, or a ketone in the case of a ketose) suffers a nucleophilic addition by one of the hydroxyls in the chain, preferably one that will form a 5 or 6 membered ring after the reaction.
As such, the anomeric carbon will have two oxygens attached (The original one and the one that bonded when the ring closed).
It will be chiral, given that it has 4 different groups attached. (-OR,-OH,-H and -R, where R is the carbon chain).
The hydroxyl group can be in any position (Above of below the ring), depending on with side the addition took place. (See attachment)
It is the carbon of the carbonyl in the open-chain form of the sugar, because it is the only one that can react with the Hydroxyls.
