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serg [7]
3 years ago
10

How many particles are in a 151 g sample of Li2O?

Chemistry
1 answer:
neonofarm [45]3 years ago
4 0

Answer:

3.052 × 10^24 particles

Explanation:

To get the number of particles (nA) in a substance, we multiply the number of moles of the substance by Avogadro's number (6.02 × 10^23)

The mass of Li2O given in this question is as follows: 151grams.

To convert this mass value to moles, we use;

moles = mass/molar mass

Molar mass of Li2O = 6.9(2) + 16

= 13.8 + 16

= 29.8g/mol

Mole = 151/29.8g

mole = 5.07moles

number of particles (nA) of Li2O = 5.07 × 6.02 × 10^23

= 30.52 × 10^23

= 3.052 × 10^24 particles.

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IF 500 mL of air contains 2.50x10^22 particles (atoms and molecules). how much particles do you inhale in one day if you breath
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I think it is 0.00192

Explanation:

28.795705

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=0.00192

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How many moles of Kl are needed to make 4.07 L of a 1.138 M solution?
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Answer:

A

Explanation:

cause

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How is a simple cell constructed​
ki77a [65]

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a simple voltaic cell is made by immersing one zinc plate and one copper plate inside water diluted sulfuric acid solution.

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Read 2 more answers
Night visin cameras are sensitive to energies around 2.21 x 10-19 J. What wavelength of electromagnetic radiation do they use?
defon

Answer:

8.99×10^-7m

Explanation:

The wavelength can be calculated using the expression below

E=hcλ

Where E= energy= 2.21 x 10^-19 J.

C= speed of light= 3x10^8 m/s

h= planks constant= 6.626 × 10^-34 m2 kg / s

E=hcλ

λ= E/(hc)

Substitute for the values

λ=( 2.21 x 10^-19 )/(6.626 × 10^-34 × 3x10^8 )

= 8.99×10^-7m

6 0
3 years ago
If the distance between two objects is decreased to - of the original
Charra [1.4K]

Answer:

The new force will be \frac{1}{100} of the original force.

Explanation:

In the context of this problem, we're dealing with the law of gravitational attraction. The law states that the gravitational force between two object is directly proportional to the product of their masses and inversely proportional to the square of a distance between them.

That said, let's say that our equation for the initial force is:

F = G\frac{m_1m_2}{R^2}The problem states  that  the distance decrease to 1/10 of the original distance, this means:[tex]R_2 = \frac{1}{10}R

And the force at this distance would be written in terms of the same equation:

F_2 = G\frac{m_1m_2}{R_2^2}

Find the ratio between the final and the initial force:

\frac{F_2}{F} = \frac{G\frac{m_1m_2}{R_2^2}}{G\frac{m_1m_2}{R^2}}

Substitute the value for the final distance in terms of the initial distance:

\frac{F_2}{F} = \frac{G\frac{m_1m_2}{(\frac{R}{10})^2}}{G\frac{m_1m_2}{R^2}}

Simplify:

\frac{F_2}{F} = \frac{\frac{1}{100R^2}}{\frac{1}{R^2}}=\frac{1}{100}

This means the new force will be \frac{1}{100} of the original force.

8 0
3 years ago
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