In the follow
<span>1)N<span>H4</span>OH,N<span>H4</span>Cl<span>O3</span>,(N<span>H4</span><span>)2</span>S<span>O3</span>,(N<span>H4</span><span>)2</span>HP<span>O4</span></span>
<span>2)Al(OH<span>)3</span>,Al(Cl<span>O3</span><span>)3</span>,A<span>l2</span>(S<span>O3</span><span>)3</span>,A<span>l2</span>(HP<span>O4</span><span>)3</span></span>
<span><span>3)Pb(OH<span>)4</span>,Pb(Cl<span>O3</span><span>)4</span>,Pb(S<span>O3</span><span>)2</span>,Pb(HP<span>O4</span><span>)2</span></span></span>
Answer:
See explanation
Explanation:
In this case, we have to remember that if we want to remove water from the reaction vessel we have to heat the vessel. So, we can convert the liquid water into <u>gas water</u> and we can remove it from the vessel. In this case, the products of dehydration for both molecules are <u>(E)-4-methylpent-2-ene</u> and <u>cyclohexene</u> with boiling points of <u>59.2 ºC</u> and <u>89 ºC</u> respectively. The boiling point of water is <u>100 ºC</u>, therefore if we heat the vessel the products and water would leave the system, and the products would be lost.
See figure 1
I hope it helps!
Answer:
1. V₁ = 2.0 mL
2. V₁ = 2.5 mL
Explanation:
<em>You are provided with a stock solution with a concentration of 1.0 × 10⁻⁵ M. You will be using this to make two standard solutions via serial dilution.</em>
To calculate the volume required (V₁) in each dilution we will use the dilution rule.
C₁ . V₁ = C₂ . V₂
where,
C are the concentrations
V are the volumes
1 refers to the initial state
2 refers to the final state
<em>1. Perform calculations to determine the volume of the 1.0 × 10⁻⁵ M stock solution needed to prepare 10.0 mL of a 2.0 × 10⁻⁶ M solution.</em>
C₁ . V₁ = C₂ . V₂
(1.0 × 10⁻⁵ M) . V₁ = (2.0 × 10⁻⁶ M) . 10.0 mL
V₁ = 2.0 mL
<em>2. Perform calculations to determine the volume of the 2.0 × 10⁻⁶ M solution needed to prepare 10.0 mL of a 5.0 × 10⁻⁷ M solution.</em>
C₁ . V₁ = C₂ . V₂
(2.0 × 10⁻⁶ M) . V₁ = (5.0 × 10⁻⁷ M) . 10.0 mL
V₁ = 2.5 mL
Answer:-
Oxygen gains electrons and is reduced.
Explanation:-
For this reaction the balanced chemical equation is
4Fe + 3O2 --> 2Fe2O3
When Oxygen is present as oxygen gas, the oxidation number of O is Zero since it is the only element present in Oxygen gas.
Similarly Iron is present in Fe with oxidation number Zero.
In the case of Fe2O3, Oxygen has the oxidation number -2 while Iron has +3.
So the oxidation number of Oxygen goes from Zero to -2.
Since the oxidation number decreases Oxygen is reduced.
Since reduction involves gain of electrons, Oxygen gains electrons.
