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3 years ago

How many moles of pentane are in 1.48x1021

Chemistry

2 answers:

vodka [1.7K]3 years ago

7 0

multiply the number of each atom with its molecular mass. (see the periodic table)

5*12 + 12*1 = 72 gram per mole.

Or more accurately: 5*12.01078 + 12*1.007947 = 72.149264 g/mole

mixer [17]3 years ago

3 0

<h2>Answer:</h2>

1.48 * 1021 number of molecules of pentane = 0.0002458 mol

Given data:

Number of molecules of pentane = 1.48 * 1021

Number of moles = ?

Solution:

1 mole = 6.02 × 1023 molecules.

1 molecule = 1/6.02 × 1023

1.48 × 1021 molecules = (1/6.02 × 1023) × 1.48 × 1021

= 2.458 × 10-3 mol

= 0.0002458 mol

So there are 0.000245 moles of pentane, if the number of molecules are 1.48 * 1021. Because 1 mole mole contain Avogadro's number of molecules.

Which is equal to 6.02 × 1023 molecules.

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What is the solution to the problem expressed to the correct number of significant figures? 2.13 + 1 = ?

Paha777 [63]

So,

With addition, we the last digit we keep will be the one which is known for both individual values.

We know 2.13 to the hundredths, but we only know 1 to the ones. Therefore, we will round off in the ones place.

2.13 + 1 = 3.13 (unrounded)

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Hope this helps!

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3 years ago

ou will prepare 250-mL of this solution using a 30% (m/v) NaOH stock solution. How many mL of the NaOH stock solution will you n

ArbitrLikvidat [17]

Answer:

$\boxed{\text{3.3 mL}}$

Explanation:

You must convert 30 % (m/v) to a molar concentration.

Assume 1 L of solution.

1. Mass of NaOH

$\text{Mass of NaOH} = \text{1000 mL solution } \times \dfrac{\text{30 g NaOH}}{\text{100 mL solution}} = \text{300 g NaOH}$

2. Moles of NaOH

$\text{Moles of NaOH} = \text{300 g NaOH} \times \dfrac{\text{1 mol NaOH}}{\text{40.00 g NaOH}} = \text{7.50 mol NaOH}$

3. Molar concentration of NaOH

$c= \dfrac{\text{moles}}{\text{litres}} = \dfrac{\text{7.50 mol}}{\text{1 L}} = \text{7.50 mol/L}$

4. Volume of NaOH

Now that you know the concentration, you can use the dilution formula .

$c_{1}V_{1} = c_{2}V_{2}$

to calculate the volume of stock solution.

Data:

c₁ = 7.50 mol·L⁻¹; V₁ = ?

c₂ = 0.1 mol·L⁻¹; V₂ = 250 mL

Calculations:

(a) Convert millilitres to litres

$V = \text{250 mL} \times \dfrac{ \text{1 L}}{\text{1000 mL}} = \text{0.250 L}$

(b) Calculate the volume of dilute solution

$\begin{array}{rcl}7.50V_{1} & = & 0.1 \times 0.250\\7.50V_{1} &= & 0.0250\\V_{1} & = & \text{0.0033 L}\\& = & \textbf{3.3 mL}\\\end{array}$

$\text{You will need $\boxed{\textbf{3.3 mL}}$ of the stock solution.}$

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