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3 years ago

Why are familiar objects such as pens and paper clips not commonly counted in moles?

2 answers:

8 0

The number is far to big. A single mole of jelly beans would take up the whole planet earth. So imagine the numbers we would have to go into!

blondinia [14]3 years ago

8 0

One mole is equal to avagadro number

Avagadro's number = 6.023 X 10^23

So it is a big big number. For pens or paper clips we do not need such big unit to express any figure.

Hence we cannot count the normal pens or pencils type macro objects in terms of moles

The mole is used for atomic level particles where in a small amount such large number of identities are present

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How many moles of Al would be produced from 20 moles of Al2O3?<br> 2Al2O3<br> -><br> 4A1 + 302

Evgesh-ka [11]

<h3>Answer:</h3>

$\displaystyle 40 \ mol \ Al$

<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>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] 2Al₂O₃ → 4Al + 3O₂

[Given] 20 mol Al₂O₃

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

[RxN] 2 mol Al₂O₃ → 4 mol Al

<u>Step 3: Stoich</u>

[DA] Set up: $\displaystyle 20 \ mol \ Al_2O_3(\frac{4 \ mol \ Al}{2 \ mol \ Al_2O_3})$
[DA] Multiply/Divide [Cancel out units]: $\displaystyle 40 \ mol \ Al$

<u>Step 4:Check</u>

<em>Follow sig fig rules and round. We are given 1 sig fig.</em>

Since our final answer already has 1 sig fig, there is no need to round.

4 0

3 years ago

A 508-g sample of sodium bicarbonate (NaHCO3) contains how many moles of sodium bicarbonate (NaHCO3)?

Blizzard [7]

molar mass = (22.99) + (1.01) + (12.01) + 3(16.00)

molar mass = 84.01 g/mol

(508g)(1 mol/84.01 g) = 6.0

There are 6.0 moles of sodium bicarbonate

3 0

3 years ago

Is this statement true or false?

Snezhnost [94]

The correct answer is False.

5 0

2 years ago

Read 2 more answers

The vapor pressure of pure water at 296 K is 2778.5 Pa. The vapor forms an ideal gas. 1) In some oil, the equilibrium concentrat

Murljashka [212]

Explanation:

It is given that vapor pressure of pure water at 296 K is 2778.5 Pa.These vapors will result in the formation of an ideal gas.

Now, as water is covered with oil and contains only 1% molecules of water. Hence, the vapor pressure of this mixture will also be equal to the vapor pressure of pure water.

So, vapor pressure of mixture = 1% vapor pressure of pure water

Therefore, $\text{(Vapor pressure)}_{mixture}$ = $\frac{1}{100} \times 2778.5 Pa$

= 27.785 Pa

Thus, we can conclude that the equilibrium vapor pressure of water above the oil layer is 27.785 Pa.

3 0

3 years ago

Pb(NO3)2+K2CrO4 --> PbCrO4+KNO3 <br> How do you balance this?

belka [17]

Answer:

Pb(NO₃)₂ + K₂CrO₄ ⟶ PbCrO₄ + 2KNO₃

Step-by-step explanation:

The unbalanced equation is

Pb(NO₃)₂ + K₂CrO₄ ⟶ PbCrO₄ + KNO₃

Notice that the complex groups like NO₃ and CrO₄ stay the same on each side of the equation.

One way to simplify the balancing is to replace them with a single letter.

(a) For example, let <em>X = NO₃</em> and <em>Y =CrO₄</em>. Then, the equation becomes

PbX₂ + K₂Y ⟶ PbY + KX

(b) You need 2X on the right, so put a 2 in front of KX.

PbX₂ + K₂Y ⟶ PbY + 2KX

Pb(NO₃)₂ + K₂CrO₄ ⟶ PbCrO₄ + 2KNO₃

5 0

3 years ago