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serg [7]
2 years ago
10

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

Chemistry
1 answer:
neonofarm [45]2 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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Now, when you ask if it can only pound one at a time, my honest answer is, it depends. Depending on the size of your mortar, you could grind materials two or three at a time. But if you are concerned with contamination, then you do it one at a time, especially if you don't want them to get mixed up.

8 0
2 years ago
a 4.50 g coin of copper absorbed 54 calories of heat. what was the final temperature of the copper if the initial temperature wa
vlada-n [284]

Answer:

Final temperature =  T₂ = 155.43 °C

Explanation:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Given data:

Mass of coin = 4.50 g

Heat absorbed = 54 cal

Initial temperature = 25 °C

Specific heat of copper = 0.092 cal/g °C

Final temperature = ?

Solution:

Q = m.c. ΔT

ΔT = T₂ -T₁

Q = m.c. T₂ -T₁

54 cal = 4.50 g × 0.092 cal/g °C ×  T₂ -25  °C

54 cal = 0.414 cal/ °C ×  T₂ -25  °C

54 cal /0.414 cal/ °C =  T₂ -25  °C

130.43 °C  =  T₂ -25 °C

130.43 °C + 25 °C = T₂

155.43 °C = T₂

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Answer:

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Explanation:

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