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2 years ago

How many atoms of carbon are in 6.00g of carbon?(Mm of carbon is 12g/mol)

Chemistry

1 answer:

xenn [34]2 years ago

5 0

<h3>Answer:</h3>

3.01 × 10²³ atoms C

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

Chemistry

Atomic Structure

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

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<h3>Explanation:</h3>

Step 1: Define

6.00 g C

Step 2: Identify Conversions

Avogadro's Number

Molar Mass of C - 12.01 g/mol

Step 3: Convert

$6.00 \ g \ C(\frac{1 \ mol \ C}{12.01 \ g \ C} )(\frac{6.022 \cdot 10^{23} \ atoms \ C}{1 \ mol \ C} )$ = 3.00849 × 10²³ atoms C

Step 4: Check

We are given 3 sig figs. Follow sig fig rules and round.

3.00849 × 10²³ atoms C ≈ 3.01 × 10²³ atoms C

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2 years ago

Student mixed 25.0 mL of 0.100 M glucose, 15.0 mL of 0.500 M NaCl and 450. mL water. What are concentrations in his solution?

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The question is incomplete, here is the complete question:

Student mixed 25.0 mL of 0.100 M glucose, 15.0 mL of 0.500 M NaCl and 450. mL water. What are concentrations in his solution?

A. 5.10 mM glucose, 15.3 mM NaCl

B. 5.56 mM glucose, 16.7 mM NaCl

C. 0.556 mM glucose, 0.167 mM NaCl

D. 0.222 mM glucose, 1.11 mM NaCl

E. 0.556 mM glucose, 0.0667 mM NaCl

Answer: The concentration of glucose and NaCl in the solution is 5.10 mM and 15.3 mM respectively.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ .....(1)

For glucose:

Molarity of glucose solution = 0.100 M

Volume of solution = 25.0 mL

Putting values in equation 1, we get:

$0.100M=\frac{\text{Moles of glucose}\times 1000}{25.0}\\\\\text{Moles of glucose}=\frac{(0.100\times 25.0)}{1000}=0.0025mol$

For NaCl:

Molarity of NaCl solution = 0.500 M

Volume of solution = 15.0 mL

Putting values in equation 1, we get:

$0.500M=\frac{\text{Moles of NaCl}\times 1000}{15.0}\\\\\text{Moles of NaCl}=\frac{(0.500\times 15.0)}{1000}=0.0075mol$

Total volume of solution = 25.0 + 15.0 + 450. = 490. mL

Now, calculating the concentration of glucose and NaCl in the solution by using equation 1:

Using conversion factor: 1 M = 1000 mM

For glucose:

Moles of glucose = 0.0025 moles

Volume of solution = 490. mL

Putting values in equation 1, we get:

$\text{Concentration of glucose}=\frac{0.0025\times 1000}{490}\\\\\text{Concentration of glucose}=0.0051M=5.10mM$

For NaCl:

Moles of NaCl = 0.0075 moles

Volume of solution = 490. mL

Putting values in equation 1, we get:

$\text{Concentration of NaCl}=\frac{0.0075\times 1000}{490}\\\\\text{Concentration of NaCl}=0.0153M=15.3mM$

Hence, the concentration of glucose and NaCl in the solution is 5.10 mM and 15.3 mM respectively.

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3 years ago