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2 years ago

0.58 mol of Mg contains how many atoms? please show work

Chemistry

1 answer:

Jet001 [13]2 years ago

3 0

The number of atoms present in 0.58 mole of magnesium, Mg is 3.49×10²³ atoms

<h3>Avogadro's hypothesis </h3>

1 mole of Mg = 6.02×10²³ atoms

<h3>How to determine the atoms in 0.58 mole of Mg </h3>

1 mole of Mg = 6.02×10²³ atoms

Therefore,

0.58 mole of Mg = 0.58 × 6.02×10²³

0.58 mole of Mg = 3.49×10²³ atoms

Thus, 3.49×10²³ atoms are present in 0.58 mole of Mg

Learn more about Avogadro's number:

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2 moles of sodium chloride are dissolved in 8 liters of water. What is its molarity (M) ?

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Answer:

0.25M

Explanation:

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Molarity= number of moles ÷ volume in liters

Given: number of moles= 2

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Which compound is an Arrhenius base?

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3 years ago

The heat capacity of water is 1cal degree'g1 (1 calorie per degree centigrade, per gram). You are given 1 gallon of water at 25

sdas [7]

Answer:

The heat needed to boil 1 gallon of water is 81,490.62 Joules.

Explanation:

$Q=mc\Delta T$

Where:

Q = heat absorbed or heat lost

c = specific heat of substance

m = Mass of the substance

ΔT = change in temperature of the substance

We have :

Volume of water = V = 1 gal = 4546.09 mL

Density of water , d= 1 g/mL

mass of water = m = d × V = 1g/mL × 4546.09 mL = 4546.09 g

Specific heat of water = c = 1 Cal/g°C

ΔT = 100°C - 25°C = 75 °C

9 (boiling pint of water is 100°C)

Heat absorbed by the water to make it boil:

$Q= 4546.09 g\times 1 Cal/g^oC\times 75^oC=340,956.75 Cal$

1 calorie = 4.184 J

$Q=\frac{340,956.75}{4.184} J = 81,490.62 J$

The heat needed to boil 1 gallon of water is 81,490.62 Joules.

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