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Angelina_Jolie [31]
3 years ago
9

Calculate the number of ammonia molecules in 3.9 g.

Chemistry
1 answer:
disa [49]3 years ago
6 0
•3.9g of ammonia
•molar mass of ammonia = 17.03g/mol

1st you have to covert grams to moles by dividing the mass of ammonia with the molar mass:

(3.9 g)/ (17.03g/mol) = 0.22900763mols

Then convert the moles to molecules by multiplying it with Avogadro’s number:

Avogadro’s number: 6.022 x 10^23


0.22900763mols x (6.022 x 10^23 molecs/mol)
= 1.38 x 10^23 molecules
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Answer:

beryllium iodide has a molar mass of 262.821 g mol−1 , which means that 1 mole of beryllium iodide has a mass of 262.821 g . To find the mass of 0.02 moles of beryllium iodide, simply multiply the number of moles by the molar mass in conversion factor form.

Explanation:

5 0
3 years ago
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A tank of 0.1m3 volume contains air at 25∘C and 101.33 kPa. The tank is connected to a compressed-air line which supplies air at
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Answer:

Amount of Energy = 23,467.9278J

Explanation:

Given

Cv = 5/2R

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The energy balance in the tank is given as

∆U = Q + W

According to the first law of thermodynamics

In the question, it can be observed that the volume of the reactor is unaltered

So, dV = W = 0.

The Internal energy to keep the tank's constant temperature is given as

∆U = Cv((45°C) - (25°C))

∆U = Cv((45 + 273) - (25 + 273))

∆U = Cv(20)

∆U = 5/2 * 8.314 * 20

∆U = 415.7 J/mol

Before calculating the heat loss of the tank, we must first calculate the amount of moles of gas that entered the tank where P1 = 101.33 kPa

The Initial mole is calculated as

(P * V)/(R * T)

Where P = P1 = 101.33kPa = 101330Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

So, n = (101330 * 0.1)/(8.314*298)

n = 4.089891232222

n = 4.089

Then we Calculate the final moles at P2 = 1500kPa = 1500000Pa

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R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

n = (1500000 * 0.1)/(8.314*298)

n = 60.54314465936812

n = 60.543

So, tue moles that entered the tank is ∆n

∆n = 60.543 - 4.089

∆n = 56.454

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Q = 415.7 * 56.454

Q = 23,467.9278J

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3 years ago
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FeBr₃ ⇒ limiting reactant

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<h3>Further explanation</h3>

Reaction

2FeBr₃ + 3Na₂S → Fe₂S₃ + 6NaBr

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So  FeBr₃ as a limiting reactant(smaller ratio)

mol NaBr based on limiting reactant (FeBr₃) :

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Water, H2O, HOH these are all the same compound just worded differently
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