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3 years ago

Suppose we have a compound that is 4.330 % Li, 22.10 % Cl, 39.89 % O, and 33.69 % H2O. What is the compounds formula?

Chemistry

1 answer:

amid [387]3 years ago

8 0

The formula of compound is LiClO4.3H2O

<em><u>calculation</u></em>

<em><u> </u></em>find the mole of each element

that is moles for Li,Cl,O and that of H2O

moles = % composition/ molar mass

For Li = 4.330/ 6.94 g/mol= 0.624 moles

Cl=22.10/35.5=0.623 moles

39.89/16 g/mol =2.493 moles

H20= 33.69/18 g/mol= 1.872 moles

find the mole ratio by dividing each moles by smallest number of mole ( 0.624 moles)

that is for Li= 0.624/0.623= 1

Cl= 0.623/0.623=1

O = 2.493/0.623 =4

H2O= 1.872/0.623=3

<h3>Therefore the formula=LiClO4.3H2O</h3><h3 />

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How many moles of water were lost if the amount of water lost was 0.456 grams? Do not include units and assume three significant

nika2105 [10]

<h3>Answer:</h3>

0.0253 mol H₂O

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 0.456 g H₂O (water)

<u>Step 2: Identify Conversions</u>

[PT] Molar Mass of H - 1.01 g/mol

[PT] Molar Mass of O - 16.00 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

<u>Step 3: Convert</u>

[DA] Set up: $\displaystyle 0.456 \ g \ H_2O(\frac{1 \ mol \ H_2O}{18.02 \ g \ H_2O})$
[DA] Multiply/Divide [Cancel out units]: $\displaystyle 0.025305 \ mol \ H_2O$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

0.025305 mol H₂O ≈ 0.0253 mol H₂O

3 0

2 years ago

Find the mass of 3.00 mol of acetic acid, C2H4O2.

Alexeev081 [22]

Molar mass

C₂H₄O₂ = 60.0 g/mol

n = mass / molar mass

3.00 = mass / 60.0

m = 3.00 * 60.0

m = 180 g of <span>C₂H₄O₂

hope this helps!</span>

5 0

2 years ago

You are a researcher for a golf club manufacturer. You are given two identical looking cubes of a metal alloy. You are informed

Rashid [163]

Answer:

C.Melt both cubes and look for a broader range of melting temperatures. The one that melts over a broader range of temperatures is the amorphous solid.

Explanation:

Amorphous solids is one that do not have a fixed melting points but melt over a wide range of temperature due to the irregular shape hence its name. Contrariwise crystalline solids, have a fixed and sharp melting point.

This comes in handy to solve the riddle. We can characterise the pair with the melting point property.

7 0

3 years ago

The first-order rate constant for the reaction of methyl chloride (CH3Cl) with water to produce methanol (CH3OH) and hydrochlori

Dvinal [7]

Answer:

K(48.5°C) = 1.017 E-8 s-1

Explanation:

CH3Cl + H2O → CH3OH + HCl

at T1 = 25°C (298 K) ⇒ K1 = 3.32 E-10 s-1

at T2 = 48.5°C (321.5 K) ⇒ K2 = ?

Arrhenius eq:

K(T) = A e∧(-Ea/RT)
Ln K = Ln(A) - [(Ea/R)(1/T)]

∴ A: frecuency factor

∴ R = 8.314 E-3 KJ/K.mol

⇒ Ln K1 = Ln(A) - [Ea/R)*(1/T1)]..........(1)

⇒ Ln K2 = Ln(A) - [(Ea/R)*(1/T2)].............(2)

(1)/(2):

⇒ Ln (K1/K2) = (Ea/R)* (1/T2-1/T1)

⇒ Ln (K1/K2) = (116 KJ/mol/8.3134 E-3 KJ/K.mol)*(1/321.5 K - 1/298 K)

⇒ Ln (K1/K2) = (13952.37 K)*(- 2.453 E-4 K-1)

⇒ Ln (K1/K2) = - 3.422

⇒ K1/K2 = e∧(-3.422)

⇒ (3.32 E-10 s-1)/K2 = 0.0326

⇒ K2 = (3.32 E-10 s-1)/0.0326

⇒ K2 = 1.017 E-8 s-1

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3 years ago

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Brums [2.3K]

This equation is impossible. NaSO4 is non-existent. Did you mean Na2SO4?

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2 years ago