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Studentka2010 [4]

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5 0

3 years ago

Read 2 more answers

How many grams of molybdenum(Mo) are in 2.68E24 atoms of Mo?

loris [4]

<h3>Answer:</h3>

427 g Mo

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

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table
Using Dimensional Analysis
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<h3>Explanation:</h3>

<u>Step 1: Define</u>

2.68E24 atoms Mo or 2.68 × 10²⁴ atoms Mo

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

Avogadro's Number

Molar Mass of Mo - 95.94 g/mol

<u>Step 3: Convert</u>

<u /> $2.68 \cdot 10^{24} \ atoms \ Mo(\frac{1 \ mol \ Mo}{6.022 \cdot 10^{23} \ atoms \ Mo} )(\frac{95.94 \ g \ Mo}{1 \ mol \ Mo} )$ = 426.966 g Mo

<u>Step 4: Check</u>

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

426.966 g Mo ≈ 427 g Mo

3 0

3 years ago

Under what conditions will deviations from the "ideal" gas be expected?

mixas84 [53]

Answer:

Conditions which results in deviating a gas from "ideal" behavior are

1. Low Temperature

2. High Temperature

Explanation:

Ideal gas according to the kinetic model theory states that the conditions that apply are high temperatures where kinetic energy and low pressure is too high and the interactions in between and the container are negligible. Hence, the deviations of ideal gas falls when there is low temperature and high pressure.

3 0

3 years ago

Calculate the moles of calcium oxalate formed from 5 grams of potassium oxalate (molar mass 184.24 g/mol) the mole mole ratio fr

Nataly_w [17]

5 g of potassium oxalate react to produce 0.03 moles of calcium oxalate.

Calcium oxalate (CaC₂O₄) is obtained by the reaction of 5 g of potassium oxalate (K₂C₂O₄).

We can calculate the moles of CaC₂O₄ obtained considering the following relationships.

The molar mass of K₂C₂O₄ is 184.24 g/mol.
The mole ratio of K₂C₂O₄ to CaC₂O₄ is 2:1.

$5 g K_2C_2O_4 \times \frac{1molK_2C_2O_4}{184.24gK_2C_2O_4} \times \frac{1molCaC_2O_4}{2molK_2C_2O_4} = 0.03molCaC_2O_4$

5 g of potassium oxalate react to produce 0.03 moles of calcium oxalate.

Learn more: brainly.com/question/15288923

7 0

3 years ago

The following reaction shows the products when sulfuric acid and aluminum hydroxide react.

SSSSS [86.1K]

Answer:

There will remain 11.47 grams of Al(OH)3

Explanation:

Step 1: Data given

Mass of sulfuric acid = 35.0 grams

Molar mass sulfuric acid = 98.08 g/mol

Mass of aluminium hydroxide = 30.0 grams

Molar mass of aluminium hydroxide = 78.0 g/mol

Step 2: The balanced equation

2Al(OH)3 + 3H2SO4 → Al2(SO4)3 + 6H2O

Step 3: Calculate moles

Moles = mass / molar mass

Moles Al(OH)3 = 30.0 grams / 78.0 g/mol

Moles Al(OH)3 = 0.385 moles

Moles H2SO4 = 35.0 grams / 98.08 g/mol

Moles H2SO4 = 0.357 moles

Step 4: Calculate the limiting reactant

For 2 moles Al(OH)3 we need 3 moles H2SO4 to produce 1 mol Al(SO4)3 and 6 moles H20

H2SO4 is the limiting reactant. It will completely be consumed ( 0.357 moles). Al(OH)3 is in excess. There will be react 2/3 * 0.357 = 0.238 moles

There will remain 0.385 - 0.238 = 0.147 moles

Mass of Al(OH)3 remaining = 0.147 moles* 78.0 g/mol = 11.47 grams

There will remain 11.47 grams of Al(OH)3

6 0

3 years ago