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3 years ago

Find the mass in grams of 0.75 moles of magnesium (Mg).

Chemistry

2 answers:

hoa [83]3 years ago

8 0

Answer:

$\boxed {\boxed {\sf 18.22875 \ g \ Mg}}$

Explanation:

We are asked to find the mass of 0.75 moles of magnesium.

Moles are converted to grams using the molar mass. This is the number of grams in 1 mole of a substance. The values are found on the Periodic Table because they are equal to the atomic masses, but the units are grams per mole instead of atomic mass units.

Look up the molar mass of magnesium.

Mg: 24.305 g/mol

We convert using dimensional analysis. Set up a conversion factor using the molar mass.

$\frac { 24.305 \ g \ Mg}{1 \ mol \ Mg}$

We are converting 0.75 moles of magnesium to grams, so we multiply the conversion factor by this value.

$0.75 \ mol \ Mg *\frac { 24.305 \ g \ Mg}{1 \ mol \ Mg}$

The units of moles of magnesium cancel.

$0.75 *\frac { 24.305 \ g \ Mg}{1 }$

$0.75 *{ 24.305 \ g \ Mg}$

$18.22875 \ g \ Mg$

0.75 moles of magnesium contains 18.22875 grams of magnesium.

Lady bird [3.3K]3 years ago

3 0

Answer:

18.22874999999973

I recommend you to round the nearest 1 d.p

Explanation:

have a great day!

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3 years ago

In a typical mass spectrum, a smaller signal is observed at a mass 1 amu higher than the molecular ion peak. Why

Nina [5.8K]

Answer:

Answer is explained below in the explanation section

Explanation:

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3 years ago

You are pushing a box with 20 N of force that has a mass of 10 kg, solve for acceleration

insens350 [35]

Answer:

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4 0

4 years ago

When there are 0.0814 moles of hockey pucks how many hockey pucks are there?

Dovator [93]

Answer:

4.9 x 10²²hockey pucks

Explanation:

Given parameters:

Number of moles of hockey = 0.0814moles

Unknown:

Number of pucks there = ?

Solution:

A mole of a substance is made up of Avogadro's number of particles.

Therefore;

1 mole of hockey pucks will contain 6.02 x 10²³ hockey pucks

0.0814 mole of hockey pucks will contain :

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8 0

3 years ago

Find the mass of 3.27 x 10^23 molecules of H2SO4. Use 3 significant digits and put the units.

marta [7]

Answer:

Approximately $53.3\; \rm g$ .

Explanation:

Lookup Avogadro's Number: $N_{\rm A} = 6.02\times 10^{23}\; \rm mol^{-1}$ (three significant figures.)

Lookup the relative atomic mass of $\rm H$ , $\rm S$ , and $\rm O$ on a modern periodic table:

$\rm H$ : $1.008$ .
$\rm S$ : $32.06$ .
$\rm O$ : $15.999$ .

(For example, the relative atomic mass of $\rm H$ is $1.008$ means that the mass of one mole of $\rm H\!$ atoms would be approximately $1.008\!$ grams on average.)

The question counted the number of $\rm H_2SO_4$ molecules without using any unit. Avogadro's Number $N_{\rm A}$ helps convert the unit of that count to moles.

Each mole of $\rm H_2SO_4$ molecules includes exactly $(1\; {\rm mol} \times N_\text{A}) \approx 6.02\times 10^{23}$ of these $\rm H_2SO_4 \!$ molecules.

$3.27 \times 10^{23}$ $\rm H_2SO_4$ molecules would correspond to $\displaystyle n = \frac{N}{N_{\rm A}} \approx \frac{3.27 \times 10^{23}}{6.02 \times 10^{23}\; \rm mol^{-1}} \approx 0.541389\; \rm mol$ of such molecules.

(Keep more significant figures than required during intermediary steps.)

The formula mass of $\rm H_2SO_4$ gives the mass of each mole of $\rm H_2SO_4\!$ molecules. The value of the formula mass could be calculated using the relative atomic mass of each element:

$\begin{aligned}& M({\rm H_2SO_4}) \\ &= (2 \times 1.008 + 32.06 + 4 \times 15.999)\; \rm g \cdot mol^{-1} \\ &= 98.702\; \rm g \cdot mol^{-1}\end{aligned}$ .

Calculate the mass of approximately $0.541389\; \rm mol$ of $\rm H_2SO_4$ :

$\begin{aligned}m &= n \cdot M \\ &\approx 0.541389\; \rm mol \times 98.702\; \rm g \cdot mol^{-1}\\ &\approx 53.3\; \rm g\end{aligned}$ .

(Rounded to three significant figures.)

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3 years ago