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Nata [24]
3 years ago
6

How many atoms of Mg are present in 97.22 grams of Mg?

Chemistry
1 answer:
Solnce55 [7]3 years ago
6 0

Answer:

2.439 x 10E24 atoms

Explanation:

from

N=n x NA where

N- number of atoms

n-number of moles and NA is the avogadro's number

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How much pressure would 0.8 moles of a gas at 370K exert if it occupied 17.3L of space
dezoksy [38]

Answer:

1.40 atm is the pressure for the gas

Explanation:

An easy problem to solve with the Ideal Gases Law:

P . V = n . R .T

T° = 370K

V = 17.3L

n = 0.8 mol

Let's replace data → P . 17.3L = 0.8mol . 0.082L.atm/mol.K . 370K

P = (0.8mol . 0.082L.atm/mol.K . 370K) / 17.3L = 1.40 atm

7 0
3 years ago
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5.943x10^24 molecules of H3PO4 will need how many grams of Mg(OH)2 in the reaction below? 3 Mg(OH)2 + 2 H3PO4 -------> 1 Mg3(
professor190 [17]

Answer:

Mass of Mg(OH)₂ required for the reaction = 863.13 g

Explanation:

3Mg(OH)₂ + 2H₃PO₄ -------> Mg₃(PO₄)₂ + 6H₂O

(5.943 x 10²⁴) molecules of H₃PO₄ is available fore reaction. Mass of Mg(OH)₂ required for reaction.

According to Avogadro's theory, 1 mole of all substances contain (6.022 × 10²³) molecules.

This can allow us find the number of moles that (5.943 x 10²⁴) molecules of H₃PO₄ represents.

1 mole = (6.022 × 10²³) molecules.

x mole = (5.943 x 10²⁴) molecules

x = (5.943 x 10²⁴) ÷ (6.022 × 10²³)

x = 9.87 moles

From the stoichiometric balance of the reaction,

2 moles of H₃PO₄ reacts with 3 moles of Mg(OH)₂

9.87 moles of H₃PO₄ will react with y moles of Mg(OH)₂

y = (3×9.87)/2 = 14.80 moles

So, 14.8 moles of Mg(OH)₂ is required for this reaction. We them convert this to mass

Mass = (number of moles) × Molar mass

Molar mass of Mg(OH)₂ = 58.3197 g/mol

Mass of Mg(OH)₂ required for the reaction

= 14.8 × 58.3197 = 863.13 g

Hope this Helps!!!

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4 years ago
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schepotkina [342]
A - albumin
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3 years ago
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A very old tree limb contains an amount of carbon-14 that is approximately 1/8 of the current atmospheric 14C levels.. Calculate
Tomtit [17]

A.       The radioactive decay equation is N = N0e^{-ln(2)*t/T }

where T is the half-life (5730 years), N0 is the number of atoms at time t = 0 and N is the number at time t.

Rewriting this as:

(N/N0) = e^{-ln(2)*t/T }

Since N = (1/8) N0 and substituting known values:

1/8 = e^{-ln(2)*t/5730}

Taking ln of both sides:

ln(1/8)= -ln(2)*t/5730

t = - 5730 * ln(1/8) / ln (2)

t = 17,190 years

The tree was cut down 17,190 years ago.

B.   N0 = 1,500,000 carbon-14 atoms

Since N = (1/8) N0

N = 187,500 carbon atoms left

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