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Snezhnost [94]
3 years ago
14

An aqueous solution of iron(II) iodide has a concentration of 0.215 molal. The percent by mass of iron(II) iodide in the solutio

n is
Chemistry
1 answer:
Basile [38]3 years ago
8 0

Answer:

6.24%

Explanation:

Molality by definition means a measurement of the number of moles of solute in solution with 1000 gm or 1Kg solvent. Notice the difference that Molarity is defined on the volume of solution and Molality on the mass of solvent.

So, An aqueous solution of iron(II) iodide has a concentration of 0.215 molal.

means 0.215 moles are present in 1 Kg of solvent.

The molar mass of Fe2I = 309.65 g / mole

mass of FeI2 = moles x molar mass

= 0.215 x 309.65

=66.57 gm

mass % of FeI2 = mass of FeI2 x 100 / total mass

= 66.57x 100 / (1000 +66.57)

= 6.24%

                           

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notka56 [123]

Answer:

The  correct answer is A

Explanation:

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3 years ago
Deterioration of buildings, bridges, and other structures through the rusting of iron costs millions of dollars a day. The actua
GrogVix [38]

The value of ∆H when 0.250kg of iron rusts is -1.846 × 10³kJ.

The rust forms when 4.85X10³ kJ of heat is released is 888.916 g.

<h3>Chemical reaction:</h3>

4 Fe + 3O2 ------ 2Fe2O3

∆H = -1.65×10³kJ

A) Given,

mass of iron = 0.250kg = 250 g

<h3>Calculation of number of moles</h3>

moles = given mass/ molar mass

= 250/ 55.85 g/mol.

= 4.476 mol

As we know that,

For the rusting of 4 moles of Fe, ∆H = -1.65×10³kJ

For the rusting of 4.476 moles of Fe ∆H required can be calculated as

-1.65×10³kJ × 4.476 mol/ 4mol

∆H required = -1.846 × 10³kJ

Now,

when 2 mol of Fe2O3 formed, ∆H = - 1.65×10³kJ

It can be said that,

-1.65×10³kJ energy released when 2 mol of Fe2O3 formed

So, -4.6 × 10³kJ energy released when 2 mol of Fe2O3 formed

= 2 × -4.6 × 10³kJ / -1.65×10³kJ

= 5.57 mol of Fe2O3 formed

Now,

mass of Fe2O3 formed = 5.57 mol × 159.59 g/mol

= 888.916 g

Thus, we calculated that the rust forms when 4.85X10³ kJ of heat is released is 888.916 g. and the value of ∆H when 0.250kg of iron rusts is -1.846 × 10³kJ.

learn more about ∆H:

brainly.com/question/24170335

#SPJ4

DISCLAIMER:

The given question is incomplete. Below is the complete question

QUESTION:

Deterioration of buildings, bridges, and other structures through the rusting of iron costs millions of dollars a day. The actual process requires water, but a simplified equation is 4Fe(s) + 3O₂(g) → 2Fe₂O₃(s) ΔH = -1.65×10³kJ

a) What is the ∆H when 0.250kg iron rusts.

(b) How much rust forms when 4.85X10³ kJ of heat is released?

7 0
1 year ago
Please help.
mojhsa [17]

Explanation: An element is represented in the form of _{A}^{Z}\textrm{X} where,

X = Symbol of the element

A = Atomic Mass of the element X

Z = Atomic Number of element X

Hence, For the element Justwondoricium,

Symbol = Jw

Atomic number = 120

Atomic Mass = 224

Now, Atomic number = Number of electrons = Number of Protons

Number of electrons = 120

Number of Protons = 120

and Atomic Mass = Number of neutrons + Number of protons

Number of neutrons can be calculated as we know atomic mass and number of neutrons, Putting the numbers we get

Number of neutrons = 224 - 120 = 104

The nearest noble gas to the element having atomic number 120 is Oganesson (Og), which has an atomic number of 118, so the next two electrons will be filled in the 8s orbital.

Electronic Configuration of Jw is [Og]8s^2

This electronic configuration lets us know about the location of the element in periodic table.

As the electron is entering the 8th shell, it belongs to the 8th period and as the last electron enters the s-orbital, it belongs to the S-block of the periodic table.

When the s-electrons are 1, it belongs to Group 1.

When the s-electrons are 2, it belongs to Group 2.

As this element has 2 s-electrons, it belongs to the Group 2.

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nikklg [1K]

Answer:

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Explanation:

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