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polet [3.4K]
2 years ago
7

What is the density of a piece of metal in g/cm3 if its mass is determined to be 42.20 g and it is in the shape of a cube, with

edge length of 2.50 cm?
Chemistry
2 answers:
kotykmax [81]2 years ago
8 0

Answer : The density of a piece of metal is 2.7008g/cm^3

Explanation :

To calculate the volume of cube, we use the formula:

V=a^3

where,

a = edge length of cube

Given :

Edge length of cube = 2.50 cm

Volume of cube = (2.50cm)^3=15.625cm^3

Given :

Mass of metal = 42.20 g

To calculate density of a substance, we use the equation:

Density=\frac{Mass}{Volume}

Putting values in above equation, we get:

\text{Density of metal}=\frac{42.20g}{15.625cm^3}=2.7008g/cm^3

Hence, the density of a piece of metal is 2.7008g/cm^3

maria [59]2 years ago
4 0
It's a cube so the volume = edge^3
Volume = 2.5^3 cm^3 = 15.625 cm^3

density = mass / volume = 42.20 / 15.625 = 2.70 You have 3 places of accuracy.

density of object = 2.70 grams / cm^3 <<<<=== answer.
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Answer:

5

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From;

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ΔE = 6.63 * 10^-34 * 3* 10^8/93.7 * 10^-9

ΔE = 21 * 10^-19 J

ΔE = -2.18 * 10^-18 J (1/nf^2 - 1/ni^2)

21 * 10^-19 J = -2.18 * 10^-18 J (1/nf^2 - 1/ni^2)

21 * 10^-19/-2.18 * 10^-18 = (1/nf^2 - 1/1^2)

-0.963 = (1/nf^2 - 1)

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0.037 = 1/nf^2

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Answer:

= 82%

Explanation:

Percentage purity is calculated by the formula;

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but; 164 g of Ca(NO3)2 = 40 g Ca

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% purity of Ca(NO3)2 = (Mass of Ca(NO3)2/ mass of the sample)× 100

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How many atoms are in 80.45 g of magnesium?
KatRina [158]

Hello!

To find the amount of atoms that are in 80.45 grams of magnesium, we will need to know Avogadro's number and the mass of one mole of magnesium.

Avogadro's number is 6.02 x 10^23 atoms, and one mole of magnesium is equal to 24.31 grams.

1. Divide by one mole of magnesium

80.45 / 24.31 = 3.309 moles (rounded to the number of sigfigs)

2. Multiply moles by Avogadro's number

3.309 x (6.02 x 10^23) = 1.99 x 10^24 (rounded to the number of sigfigs)

Therefore, there are 1.99 x 10^24 atoms in 80.45 grams of magnesium.

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