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2 years ago

A gas cylinder contains exactly 1 mole of oxygen gas (O2). How many molecules of oxygen are in the cylinder?

Chemistry

1 answer:

Aloiza [94]2 years ago

8 0

The number of molecules of oxygen in the cylinder is 6.02×1023 molecules

Data obtained from the question

Number of mole of oxygen = 1 mole

Number of molecule of oxygen =?

From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02×1023 molecules. This implies that 1 mole of oxygen contains 6.02×1023 molecules.

Since the cylinder contains 1 mole of oxygen, then the number of molecules of oxygen present in the cylinder is 6.02×1023 molecules.

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What's the Independent variable, The dependent variable, The constant for my project "Does Nail Polish affect the growth of nail

bonufazy [111]

Your independent variable is - The kind of nail polish used
Your dependent variable is - How much the nails have grown
And your constant is - Amount of time allowed for growth

6 0

3 years ago

You want to prepare 500.0 mL of 1.000 M KNO3 at 20°C, but the lab (and water) temperature is 24°C at the time of preparation. Ho

timama [110]

Explanation:

As per the given data, at a higher temperature, at $24^{o}C$ , the solution will occupy a larger volume than at $20^{o}C$ .

Since, density is mass divided by volume and it will decrease at higher temperature.

Also, concentration is number of moles divided by volume and it decreases at higher temperature.

At $20^{o}C$ , density of water=0.9982071 g/ml

Therefore, $\frac{concentration}{density}$ will be calculated as follows.

= $\frac{C_{1}}{d_{1}}$

= $\frac{1.000 mol/L}{0.9982071 g/ml}$

= 1.0017961 mol/g

At $24^{o}C$ , density of water = 0.9972995 g/ml

Since, $\frac{concentration}{density}$ = $\frac{C_{2}}{d_{2}}$

= $\frac{C_{2}}{0.9972995}$

Also, $\frac{C_{1}}{d_{1}}$ = $\frac{C_{2}}{d_{2}}$

so, 1.0017961 mol/g = $\frac{C_{2}}{0.9972995}$

$C_{2} = 1.0017961 \times 0.9972995$

= 0.9990907 mol/L

Therefore, in 500 ml, concentration of $KNO_{3}$ present is calculated as follows.

$C_{2}$ = $\frac{concentration}{volume}$

0.9990907 mol/L = $\frac{concentration}{0.5 L}$

concentration = 0.49954537 mol

Hence, mass (m'') = $0.49954537 mol \times 101.1032 g/mol$ = 50.5056 g (as molar mass of $KNO_{3}$ = 101.1032 g/mol).

Any object displaces air, so the apparent mass is somewhat reduced, which requires buoyancy correction.

Hence, using Buoyancy correction as follows,

m = $m''' \times \frac{(1 - \frac{d_{air}}{d_{weights}})}{(1 - \frac{d_{air}}{d})}}$

where, $d_{air}$ = density of air = 0.0012 g/ml

$d_{weight}$ = density of callibration weights = 8.0g/ml

d = density of weighed object

Hence, the true mass will be calculated as follows.

True mass(m) = $50.5056 \times \frac{(1 - \frac{0.0012}{8.0})}{(1 - (\frac{0.0012}{2.109})}$

true mass(m) = 50.5268 g

= 50.53 g (approx)

Thus, we can conclude that 50.53 g apparent mass of $KNO_{3}$ needs to be measured.

8 0

3 years ago

What is the answer please

lorasvet [3.4K]

Answer:

1 is

carbon dioxide co2

Explanation:

4 0

3 years ago

How many moles of NO2 are equivlent to 74.3 grams of NO2N=14.01/ o=16.00?

lesantik [10]

The molecular mass of $HNO_2$ is: 1·1 + 1·14.01 + 2·16 = 47.01 g/mol.

Now we can convert the mass into mol of $HNO_2$ :

$74.3\ g\ HNO_2\cdot \frac{1\ mol}{47\ g} = \bf 1,58\ mol\ HNO_2$

8 0

3 years ago

3. You measure a cube and determine that its sides are 0.65m. You place the cube on a mass scale and determine that this cube ha

amid [387]

The density of the cube in-

Kg/m³= 38.234 kg/m³

g/mL= 0.038234 g/mL

Explanation:

We know that volume of the cube equals m ³

Where “m” equals side of the cube

Given data-

Side of the cube=0.65m

mass of the cube= 10,500 gm

We know that 1000gm= 1 kg

Hence, 10,500 gm= 10.5 kg

Volume of the cube= (0.65)³

∴ Volume= 0.274625 m ³

We know that density = mass/volume

⇒Substituting the value of mass and volume, we get-

⇒Density= 10.5/0.274625= 38.234 kg/m³

We know that 1 kg/m³= 0.001 g/mL

Hence 38.234 kg/m³ would equal 0.038234 g/mL

Hence the density of the cube is 38.234 kg/m³ and 0.038234 g/mL

7 0

3 years ago