Answer:
0,051g of O₂
Explanation:
The reaction of precipitation of Fe is:
4Fe(OH)⁺(aq) + 4OH⁻(aq) + O₂(g) + 2H₂O(l) → 4Fe(OH)₃(s)
<em>-Where the Fe(OH)⁺ is Fe(II) and Fe(OH)₃ is Fe(III)-</em>
This reaction is showing you need 1 mol of O₂(g) per 4 moles of Fe(II) for a complete reaction.
85mL= 0,085L of 0,075M Fe(II) are:
0,085L*0,075M = 6,34x10⁻³ moles of Fe(II)
For a complete reaction of 6,34x10⁻³ moles of Fe(II) you need:
6,34x10⁻³ moles of Fe(II)×
=
1.59x10⁻³ moles of O₂. In grams:
1.59x10⁻³ moles of O₂×
= <em>0,051g of O₂</em>
<em></em>
I hope it helps!
Answer:
Desert
Explanation:
The adaptation shown by the given plants and animals shows that they will adapted to the desert biome.
It is so because, due to high temperature of desert some desert animals like camel have the storage of fat in humps or tails; some animals have large ears such as Jackrabbits, it helps to release body heat and adapt in high temperature; plants have thick water holding tissues to reduce water loss in heat and waxy coating that keeps the plants cooler and reduce moisture loss.
Hence, the correct answer is "Desert".
Answer:
1. HBr>HCl> H2S >BH3
2.K_a1 very large — H2SO4
K_a1= 1.7 x 10^−2 — H2SO3
K_a1 = 1.7 x 10^−7 — H2S
Explanation:
As one goes down a row in the Periodic Table the properties that determine the acid strength can be observed.
The atoms get larger in radius meaning that in strength, the strength of the bonds get weaker, conversely meaning that the acids get stronger.
For the halogen-containing acids above following the rows and periods, HBr has the strongest bond and is the strongest acid and others follow in this order.
HBr>HCl> H2S >BH3
Acid Dissociation Constant provides us with information known as the ionization constant which comes in handy to measure the acid's strength. The meaning of the proportions are thus, the higher the Ka value, the stronger the acid i.e. it liberates more number of hydrogen ions per mole of acid in solution.
In solution strong acids completely dissociate hence, the value of dissociation constant of strong acids is very high.
Following the cues above on Ka;
K_a1 very large — H2SO4
K_a1= 1.7 x 10^−2 — H2SO3
K_a1 = 1.7 x 10^−7 — H2S
Answer:
pH =3.8
Explanation:
Lets call the monoprotic weak acid HA, the dissociation equilibria in water will be:
HA + H₂O ⇄ H₃O⁺ + A⁻ with Ka = [ H₃O⁺] x [A⁻]/ [HA]
The pH is the negative log of the H₃O⁺ concentration, we know the equilibrium constant, Ka and the original acid concentration. So we will need to find the [H₃O⁺] to solve this question.
In order to do that lets set up the ICE table helper which accounts for the species at equilibrium:
HA H₃O⁺ A⁻
Initial, M 0.40 0 0
Change , M -x +x +x
Equilibrium, M 0.40 - x x x
Lets express these concentrations in terms of the equilibrium constant:
Ka = x² / (0.40 - x )
Now the equilibrium constant is so small ( very little dissociation of HA ) that is safe to approximate 0.40 - x to 0.40,
7.3 x 10⁻⁶ = x² / 0.40 ⇒ x = √( 7.3 x 10⁻⁶ x 0.40 ) = 1.71 x 10⁻³
[H₃O⁺] = 1.71 x 10⁻³
Indeed 1.71 x 10⁻³ is small compared to 0.40 (0.4 %). To be a good approximation our value should be less or equal to 5 %.
pH = - log ( 1.71 x 10⁻³ ) = 3.8
Note: when the aprroximation is greater than 5 % we will need to solve the resulting quadratic equation.
