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kondor19780726 [428]

2 years ago

8

You can use mole ratios and molar mass to determine the masses of substances involved in a chemical reaction. Using the periodic

table, calculate the molar masses of iron oxide (Fe2O3) and aluminum oxide (Al2O3). Give your answer to four significant figures.
Molar mass of Fe2O3 =
g/mol

2 answers:

Cerrena [4.2K]2 years ago

7 0

Explanation:

The molar mass of a substance is the mass in grams of one mole of the substance.

When dealing with compounds the molar mass is the sum of the of its component atomic masses and expresses in grams per mole;

Given compounds;

Fe₂O₃ and Al₂O₃

Atomic mass;

Fe = 55.845g/mol

Al = 26.9185g/mol

O = 15.9994g/mol

Molar mass of Fe₂O₃ = 2(55.845) + 3 (15.9994) =

111.69 + 47.9982 = 159.7g/mol

Molar mass of Al₂O₃ = 2(26.9185) + 3(15.9994) = 53.837‬+47.9982

= 101.8g/mol

learn more:

Molar mass brainly.com/question/2861244

#learnwithBrainly

Gennadij [26K]2 years ago

6 0

Answer:

1st one is 159.7

2nd is 102.0

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The answer is 6.1*10^-3 atm.

The pictures and explanations are there.

3 0

2 years ago

Bismuth(III) sulfate is an ionic compound formed from Bi3+ and SO42-. What is the correct way to represent the formula?

Hitman42 [59]

Answer:

Bi2(SO4)3

Explanation:

Bismuth(iii) sulfate is an ionic compound therefore, their is transfer of electron. Ionic compound has both cations and anions. The cations is positively charged ion while the anions is negatively charged ions. The cations loses electron to become positively charged while the anions gains electron to become negatively charged.

From the compound above, Bismuth(iii) sulfate the cations will be Bismuth ion which loses 3 electrons. The anions is the sulfate ion (S04)2- with a -2 charge.

The chemical formula can be computed from the charge configuration as follows

Bi3+ and (SO4)2-

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Bi2(SO4)3

Note the final chemical formula, the numbers are sub scripted

4 0

2 years ago

A standard backpack is approximately 30cm x 30cm x 40cm. Suppose you find a hoard of pure gold while treasure hunting in the wil

Blizzard [7]

Explanation:

The dimensions of a standard backpack is 30cm x 30cm x 40cm

The mass of an average student is 70 kg

We know that, the density of gold is 19.3 g/cm³.

Let m be the mass of the backpack. So,

$\text{density}=\dfrac{\text{mass}}{\text{volume}}\\\\m=d\times V\\\\m=19.3\ g/cm^3\times (30\times 30\times 40)\ cm^3\\\\m=694800\ g\\\\\text{or}\\\\m=694.8\ kg\approx 700\ kg$

An average student has a mass of 70 kg. If we compare the mass of student and mass of backpack, we find that the backpack is 10 times of the mass of the student.

8 0

2 years ago

Two substances in a mixture differ in density and particle size. These properties can be used to

lianna [129]

Answer:

(1) separate the substances

(2) chemically combine the substances

(3) determine the freezing point of the mixture

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Explanation:

3 0

2 years ago

An unknown salt is either NaF, NaCl, or NaOCl. When 0.050 mol of the salt is dissolved in water to form 0.500 L of solution, the

victus00 [196]

<u>Answer:</u> The unknown salt is NaF

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Moles of salt = 0.050 moles

Volume of solution = 0.500 L

Putting values in above equation, we get:

$\text{Molarity of salt}=\frac{0.050mol}{0.500L}\\\\\text{Molarity of salt}=0.1M$

To calculate the hydroxide ion concentration, we first calculate pOH of the solution, which is:

pH + pOH = 14

We are given:

pH = 8.08

$pOH=14-8.08=5.92$

To calculate pOH of the solution, we use the equation:

$pOH=-\log[OH^-]$

Putting values in above equation, we get:

$5.92=-\log[OH^-]$

$[OH^-]=10^{-5.92}=1.202\times 10^{-6}M$

The unknown salt given are formed by the combination of weak acid and strong acid which is NaOH

The chemical equation for the hydrolysis of $X^-$ ions follows:

$X^-(aq.)+H_2O(l)\rightleftharpoons HX(aq.)+OH^-(aq.);K_b$

<u>Initial:</u> 0.1

<u>At eqllm:</u> 0.1-x x x

Concentration of $OH^-=x=1.202\times 10^{-6}M$

The expression of $K_b$ for above equation follows:

$K_b=\frac{[OH^-][HX]}{[X^-]}$

Putting values in above expression, we get:

$K_b=\frac{(1.202\times 10^{-6})\times (1.202\times 10^{-6})}{(1-(1.202\times 10^{-6}))}\\\\K_b=1.445\times 10^{-11}M$

To calculate the acid dissociation constant for the given base dissociation constant, we use the equation:

$K_w=K_b\times K_a$

where,

$K_w$ = Ionic product of water = $10^{-14}$

$K_a$ = Acid dissociation constant

$K_b$ = Base dissociation constant = $1.445\times 10^{-11}$

Putting values in above equation, we get:

$10^{-14}=1.445\times 10^{-11}\times K_a\\\\K_a=\frac{10^{-14}}{1.445\times 10^{-11}}=6.92\times 10^{-4}$

We know that:

$K_a\text{ for HF}=6.8\times 10^{-6}$

$K_a\text{ for HCl}=1.3\times 10^{6}$

$K_a\text{ for HClO}=3.0\times 10^{-8}$

So, the calculated $K_a$ is approximately equal to the $K_a$ of HF

Hence, the unknown salt is NaF

6 0

2 years ago