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3 years ago

What is the molality of an aqueous solution containing FeCl3 (MM = 162.2 g/mol) with a mole fraction of FeCl3 of 0.15?

Chemistry

1 answer:

Natalka [10]3 years ago

3 0

Answer:

10 m

Explanation:

The mole fraction of FeCl₃ of 0.15, that is, per mole of solution, there are 0.15 moles of FeCl₃ and 1 - 0.15 = 0.85 moles of water.

The molar mass of water is 18.02 g/mol. The mass corresponding to 0.85 moles is:

0.85 mol × 18.02 g/mol = 15 g = 0.015 kg

The molality of FeCl₃ is:

m = moles of solute / kilogram of solvent

m = 0.15 mol / 0.015 kg

m = 10 m

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What do these elements make: neodymium, rhenium and sulfur

LiRa [457]

Answer:

Neodymium makes flints inside of lighters. It also takes away the green color off of glass. Rhenium is added with tungsten and molybdenum which makes filaments for lamps. Sulfur makes sulfuric acid which makes batteries and cleaners and can process ores.

Explanation:

An interesting fact for each element:

Neodymium is known for the atomic symbol Nd and the atomic number 60.

Rhenium is known for the atomic symbol Re and the atomic number 75.

Sulfur is known for the atomic symbol S and the atomic number 16.

I hope the Answer answers your question. I gave you some facts to remember those elements in the periodic table!

6 0

2 years ago

How would you prepare 3.5 L of a 0.9M solution of KCl?

Fudgin [204]

Calculate the mass of the solute in the solution :

Molar mass KCl = 74.55 g/mol

m = Molarity * molar mass * volume

m = 0.9 * 74.55 * 3.5

m = 234.8325 g

To prepare 0.9 M KCl solution, weigh 234.8325 g of salt in an analytical balance, dissolve in a beaker, shortly after transfer with the help of a funnel of transfer to a volumetric flask of 100 cm³ and complete with water up to the mark, then cover the balloon and finally shake the solution to mix

hope this helps!

8 0

3 years ago

Calculate the pH of a 0.10 M HCN solution that is 0.0070% ionized.

Anastaziya [24]

Answer:

D) 5.15

Explanation:

Step 1: Write the equation for the dissociation of HCN

HCN(aq) ⇄ H⁺(aq) + CN⁻(aq)

Step 2: Calculate [H⁺] at equilibrium

The percent of ionization (α%) is equal to the concentration of one ion at the equilibrium divided by the initial concentration of the acid times 100%.

α% = [H⁺]eq / [HCN]₀ × 100%

[H⁺]eq = α%/100% × [HCN]₀

[H⁺]eq = 0.0070%/100% × 0.10 M

[H⁺]eq = 7.0 × 10⁻⁶ M

Step 3: Calculate the pH

pH = -log [H⁺] = -log 7.0 × 10⁻⁶ = 5.15

7 0

3 years ago

Sulfur dioxide and oxygen react to form sulfur trioxide during one of the key steps in sulfuric acid synthesis. An industrial ch

Salsk061 [2.6K]

Answer:

kp= 3.1 x 10^(-2)

Explanation:

To solve this problem we have to write down the reaction and use the ICE table for pressures:

2SO2 + O2 ⇄ 2SO3

Initial 3.4 atm 1.3 atm 0 atm

Change -2x - x + 2x

Equilibrium 3.4 - 2x 1.3 -x 0.52 atm

In order to know the x value:

2x = 0.52

x=(0.52)/2= 0.26

2SO2 + O2 ⇄ 2SO3

Equilibrium 3.4 - 0.52 1.3 - 0.26 0.52 atm

Equilibrium 2.88 atm 1.04 atm 0.52 atm

with the partial pressure in the equilibrium, we can obtain Kp.

$Kp=\frac{PSO3^2}{PSO2^2 PO2}=\frac{(0.52)^2}{(2.88)^2(1.04)}=0.03135$

8 0

3 years ago

For the following reaction, 2 SO2(g) + O:(g) 2 SO,(g) the equilibrium constant, K, is 1.32 at 627°C. What is the equilibrium con

cupoosta [38]

Answer:

1.15

Explanation:

2SO₂ + O₂ ⟶ 2SO₃; K =1.32

SO₂ + ½O₂ ⟶ SO₃; K₁ = ?

When you divide an equation by 2, you take the square root of its equilibrium constant.

K₁ = √1.32 = 1.15

The equilibrium constant is 1.15.

8 0

3 years ago