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2 years ago

How many atoms are present in a sample of Calcium (Ca) weighing 127.38g?

Chemistry

1 answer:

kari74 [83]2 years ago

3 0

Answer:

1.9139 × 10²⁴ atoms Ca

General Formulas and Concepts:

Chemistry - Atomic Structure

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

Explanation:

Step 1: Define

127.38 g Ca

Step 2: Identify Conversions

Avogadro's Number

Molar Mass of Ca - 40.08 g/mol

Step 3: Convert

$127.38 \ g \ Ca(\frac{1 \ mol \ Ca}{40.08 \ g \ Ca} )(\frac{6.022 \cdot 10^{23} atoms \ Ca}{1 \ mol \ Ca} )$ = 1.91388 × 10²⁴ atoms Ca

Step 4: Check

We are given 5 sig figs. Follow sig fig rules and round.

1.91388 × 10²⁴ atoms Ca ≈ 1.9139 × 10²⁴ atoms Ca

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Whats the si unit chemist use to measure the amount of an element substance?

Westkost [7]

The SI unit for the amount of substance present is the mole.

The mole is defined as the amount of substance that has the same amount of particles as there are atoms in 12 grams of carbon-12. Mathematically, the moles of a substance may be computed using:

moles present = mass of substance / molecular mass of substance

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2 years ago

Please help with this question!

Alex17521 [72]

Answer:

A) potential energy is stored energy. Kenetic energy is energy of motion.

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2 years ago

Eading and

4vir4ik [10]

Answer: 25.8 g of $Cl_2$ will be produced from the decomposition of 73.4 g of $AuCl_3$

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} AuCl_3=\frac{73.4g}{303g/mol}=0.242moles$

The balanced chemical reaction is:

$2AuCl_3\rightarrow 2Au+3Cl_2$

According to stoichiometry :

2 moles of $AuCl_3$ produce = 3 moles of $Cl_2$

Thus 0.242 moles of will produce= $\frac{3}{2}\times 0.242=0.363mol$ of $Cl_2$

Mass of $Cl_2$ = $moles\times {\text {Molar mass}}=0.363mol\times 71g/mol=25.8g$

Thus 25.8 g of $Cl_2$ will be produced from the decomposition of 73.4 g of $AuCl_3$

5 0

2 years ago

The theoretical yield of a reaction is the amount of product obtained if the limiting reactant is completely converted to produc

Elis [28]

Answer: 9.9 grams

Explanation:

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

a) moles of $H_2$

$\text{Number of moles}=\frac{8.150g}{2g/mol}=4.08moles$

b) moles of $C_2H_4$

$\text{Number of moles}=\frac{9.330g}{28g/mol}=0.33moles$

$H_2(g)+C_2H_4(g)\rightarrow C_2H_6(g)$

According to stoichiometry :

1 mole of $C_2H_4$ combine with 1 mole of $H_2$

Thus 0.33 mole of $C_2H_4$ will combine with = $\frac{1}{1}\times 0.33=0.33$ mole of $H_2$

Thus $C_2H_4$ is the limiting reagent as it limits the formation of product.

As 1 mole of $C_2H_4$ give = 1 mole of $C_2H_6$

Thus 0.33 moles of $C_2H_4$ give = $\frac{1}{1}\times 0.33=0.33moles$ of $C_2H_6$

Mass of $C_2H_6=moles\times {\text {Molar mass}}=0.33moles\times 30g/mol=9.9g$

Thus theoretical yield (g) of $C_2H_6$ produced by the reaction is 9.9 grams

7 0

3 years ago

2Mg + O2 → 2MgO If you are burning 5.8332 g of Mg, how many grams of MgO will this make?

LekaFEV [45]

<h3>Answer:</h3>

9.6724 g MgO

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

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Chemistry

Stoichiometry

Reading a Periodic Table
Using Dimensional Analysis

<h3>Explanation:</h3>

Step 1: Define

[RxN - Balanced] 2Mg + O₂ → 2MgO

[Given] 5.8332 g Mg

Step 2: Identify Conversions

[RxN] 2 mol Mg = 2 mol MgO

Molar Mass of Mg - 24.31 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of MgO - 24.31 + 16.00 = 40.31 g/mol

Step 3: Stoichiometry

Set up: $\displaystyle 5.8332 \ g \ Mg(\frac{1 \ mol \ Mg}{24.31 \ g \ Mg})(\frac{2 \ mol \ MgO}{2 \ mol \ Mg})(\frac{40.31 \ g \ MgO}{1 \ mol \ MgO})$
Multiply/Divide: $\displaystyle 9.67241 \ g \ MgO$

Step 4: Check

Follow sig fig rules and round. We are given 5 sig figs.

9.67241 g MgO ≈ 9.6724 g MgO

5 0

2 years ago

Read 2 more answers