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nadya68 [22]
3 years ago
15

3. The density of mercury is 13.6 g/ml. Find thevolume of 150 g​

Chemistry
1 answer:
Nina [5.8K]3 years ago
6 0

Answer:

v=11.03cm

Explanation:

D=m/V

v=mass/density

150g/13.6g/ml

v=11.03cm³

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How many moles are in 2.98x10^23 particles?
Novosadov [1.4K]

Answer:

\boxed {\boxed {\sf 0.495 \ mol}}

Explanation:

We are given a number of particles and asked to convert to moles.

<h3>1. Convert Particles to Moles </h3>

1 mole of any substance contains the same number of particles (atoms, molecules, formula units) : 6.022 *10²³ or Avogadro's Number. For this question, the particles are not specified.

So, we know that 1 mole of this substance contains 6.022 *10²³ particles. Let's set up a ratio.

\frac { 1 \ mol }{6.022*10^{23 } \ particles}}

We are converting 2.98*10²³ particles to moles, so we multiply the ratio by that value.

2.98*10^{23} \ particles *\frac { 1 \ mol }{6.022*10^{23 } \ particles}}

The units of particles cancel.

2.98*10^{23}  *\frac { 1 \ mol }{6.022*10^{23 } }}

\frac { 2.98*10^{23}}{6.022*10^{23 } }}  \ mol

0.4948522086 \ mol

<h3>2. Round</h3>

The original measurement of particles (2.98*10²³) has 3 significant figures, so our answer must have the same.

For the number we found, 3 sig figs is the thousandth place.

The 8 in the ten-thousandth place (0.4948522086) tells us to round the 4 up to a 5 in the thousandth place.

0.495 \ mol

2.98*10²³ particles are equal to approximately <u>0.495 moles.</u>

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