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2 years ago

PLEASE TELL ME IF THIS IS RIGHT CLASS IN % MINUTES

Chemistry

2 answers:

alexdok [17]2 years ago

7 0

Answer:

I think that is pretty good and if you get it wrong you showed your work so your teacher can see you're thinking. That is defenetly gonna get you a score.

Explanation:

jenyasd209 [6]2 years ago

3 0

Answer:

I think is pretty good the most important thing is that you show youre work.Good luck :)

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The diameter of the He He atom is approximately 0.10 nm nm . Calculate the density of the He atom in g/cm 3 g/cm3 (assuming that

Sladkaya [172]

Answer:

Density of the He atom = 12.69 g/cm³

Explanation:

From the information given:

Since 1 mole of an atom = 6.022x 10²³ atoms)

1 atom of He = $1 \ atom \times (\dfrac{1 \ mole}{ 6.022 \times 10^{23} \ atoms}) \times ( \dfrac{4.003 \ grams}{ 1 \ mole})$

$=6.647 \times 10^{-24} \ grams$

The volume can be determined as folows:

since the diameter of the He atom is approximately 0.10 nm

the radius of the He = $\dfrac{0.10}{2}$ = 0.05 nm

Converting it into cm, we have:

$0.05 nm \times \dfrac{10^{-9} \ meters}{ 1 nm} \times \dfrac{ 1 cm }{10^{-2} \ meters}$

$=5 \times 10^{-9} \ cm$

Assuming that it is a sphere, the volume of a sphere is

= $\dfrac{4}{3}\pi r^3$

= $\dfrac{4}{3}\pi \times (5\times 10^{-9})^3$

= $5.236 \times 10^{-25} \ cm^3$

Finally, the density can be calcuated by using the formula :

$Density = \dfrac{mass}{volume}$

$D = \dfrac{6.647 \times 10^{-24} \ grams }{ 5.236 \times 10^{-25} \ cm^3}$

D = 12.69 g/cm³

Density of the He atom = 12.69 g/cm³

4 0

2 years ago

Read 2 more answers

I2 is an example of a

Free_Kalibri [48]

Answer:

number

Explanation:

5 0

3 years ago

Read 2 more answers

I need help with this question ASAP

sladkih [1.3K]

the answer is thermal.......

4 0

2 years ago

A 7.47 g sample of calcium carbonate [CaCO3 (s)] absorbs 85 J of heat, upon which the temperature of the sample increases from 2

Alex_Xolod [135]

Answer:

Specific heat of calcium carbonate(C) = 0.82 (Approx)

Explanation:

Given:

Energy absorbs (q) = 85 J

Change in temperature (Δt) = 34.9 - 21 = 13.9°C

Mass of calcium carbonate = 7.47 g

Find:

Specific heat of calcium carbonate(C)

Computation:

Specific heat of calcium carbonate(C) = q / m(Δt)

Specific heat of calcium carbonate(C) = 85 / (7.47)(13.9)

Specific heat of calcium carbonate(C) = 85 / 103.833

Specific heat of calcium carbonate(C) = 0.8186

Specific heat of calcium carbonate(C) = 0.82 (Approx)

7 0

2 years ago

How many O2 molecules react with 47 CH4 molecules according to the preceding equation CH4 + O2 → CO2 + H2O

natka813 [3]

I think 93.71 mol.
You need to use stochio

8 0

3 years ago