How many unpaired electron are there is CO2+ ion

Fe(II) can be precipitated from a slightly basic aqueous solution by bubbling oxygen through the solution, which converts Fe(II)

vovangra [49]

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)

-Where the Fe(OH)⁺ is Fe(II) and Fe(OH)₃ is Fe(III)-

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)× $\frac{1molO_2}{4moles Fe(II)}$ =

1.59x10⁻³ moles of O₂. In grams:

1.59x10⁻³ moles of O₂× $\frac{32g}{1molO_2}$ = 0,051g of O₂

I hope it helps!

7 0

3 years ago

Consider the reaction below 2SO2(g) + O2(g) ⇌ 2SO3(g) At 1000 K the equilibrium pressures of the three gases in one mixture were

Vsevolod [243]

Answer: $K_p$ for the reaction is 3.45

Explanation:

The balanced chemical reaction is:

$2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)$

At eqm. 0.562 atm 0.101 atm 0.332 atm

As we are given that:

The expression of $K_p$ for above equation follows:

$K_p=\frac{(p_{SO_3})^2}{(p_{SO_2})^2\times p_{O_2}}$

Putting values in above equation, we get:

$K_p=\frac{(0.332)^2}{(0.562)^2\times 0.101}=3.45$

The value of $K_p$ for the reaction is 3.45

8 0

3 years ago

What is the volume occupied by 6.0 moles of carbon dioxide gas at standard conditions of temperature and pressure

aliina [53]

Answer: 134.4 L

Explanation:

it just is

3 0

4 years ago

if the reduced row echelon form of the augmented matrix for a linear system has a row of zeros, then the system must have infini

BigorU [14]

Answer:

False

Explanation:

If the row echelon form of the augmented matrix for a linear system has a row of zeros, then the there must not have infinitely many solution,

we can prove this with an example. Suppose we have an augmented matrix A for linear system with a row of zeros,

1 0 0 1

A= 0 1 0 -2

0 0 1 - 1

0 0 0 0

we get

x1=1

x2=-2

x3=-1

so, system has an unique solution.

we can take inference that the given statement is wrong

7 0

3 years ago

What is the pH at the equivalence point in the titration of a 25.7 mL sample of a 0.370 M aqueous nitrous acid solution with a 0

expeople1 [14]

Answer:

pH = 8.24

Explanation:

Nitrous acid (HNO₂) reacts with KOH, thus:

HNO₂ + KOH → KNO₂ + H₂O

Moles of HNO₂ are:

0.0257mL ₓ (0.370mol / L) = 0.00951moles.

In equivalence point, the complete moles of nitrous acid reacts with KOH producing potassium nitrite. There are needed:

0.00951mol ₓ (1L / 0.491mol) = 0.01937L ≡ 19.4mL of 0.491M KOH to reach equivalence point.

Total volume in equivalence point is: 19.4mL + 25.7mL = 45.1mL

Potassium nitrite is in equilibrium with water, thus:

NO₂⁻ + H₂O ⇄ HNO₂ + OH⁻

Where equilibrium constant, Kb, is defined as:

Kb = 1.41x10⁻¹¹ = $\frac{[OH^-][HNO_2]}{[NO_2]}$

In equilibrium, molarity of each compound are:

[NO₂⁻]: 0.00951mol/0.00451L - X = 0.211M - X

[HNO₂]: X

[OH⁻]: X

Where X is reaction coordinate

Replacing in Kb:

1.41x10⁻¹¹ = $\frac{[X][X]}{[0.211 -X]}$

0 = X² + 1.41x10⁻¹¹X - 2.97x10⁻¹²

Solving for X:

X = -1.72x10⁻⁶ FALSE ANSWER. There is no negative concentrations.

X = 1.72x10⁻⁶. Right answer.

That means:

[OH⁻]: 1.72x10⁻⁶M

As pOH is -log [OH⁻] and pH = 14-pOH:

pOH = 5.76; pH = 8.24

3 0

3 years ago