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3 years ago

What amount (moles) of compound is present in 1.00 g of each of the compounds in Exercise 54?

Chemistry

1 answer:

Anika [276]3 years ago

5 0

Answer:

1. Caffeine, C₈H₁₀N₄O₂

Amount = 1.00/194 = 0.00515 moles

2. Ethanol, C₂H₅OH

Amount = 0.0217 moles

3. Dry Ice, CO₂

amount = 0.0227 moles

Note: The question is incomplete. The compound are as follows:

1. Caffeine, C₈H₁₀N₄O₂;

2. Ethanol, C₂H₅OH;

3. Dry Ice, CO₂

Explanation:

Amount (moles) = mass in grams /molar mass in grams per mole

1. Caffeine, C₈H₁₀N₄O₂

molar mass of caffeine = 194 g/mol

Amount = 1.00 g/194 g/mol = 0.00515 moles

2. Ethanol, C₂H₅OH

molar mass of ethanol = 46 g/mol

Amount = 1.00 g/46 g/mol = 0.0217 moles

3. Dry Ice, CO₂

molar mass of dry ice = 44 g/mol

amount = 1.00 g/44 g/mol = 0.0227 moles

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3 years ago

What type of reaction provide energy within a red giant star? and how come?

Dafna1 [17]

"A red giant is a star that has exhausted the primary supply of hydrogen fuel at its core. An average-sized star like our Sun will spend the final 10 percent of its life as a red giant. In this phase, a star's surface temperature drops to between 3,140 and 6,741°F (1,727 and 3,727°C) and its diameter expands to 10 to 1,000 times that of the Sun. The star takes on a reddish color, which is what gives it its name."

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3 years ago

Which statement is true? Postively charged objects attract other positively charged objects negatively charged objects attract o

ANEK [815]

Answer:

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3 years ago

What is the volume of ammonia produced at 243 K at a pressure of 1.38 atm by the unbalanced reaction on the left if 5740 moles o

Neporo4naja [7]

Answer:

49671 L is the produced volume of ammonia

Explanation:

We think the reaction of ammonia 's production:

N₂(g) + 3H₂(g) → 2NH₃ (g)

We have the moles of each reactant so let's determine the limiting reactant:

Ratio is 1:3. 1 mol of nitrogen reacts with 3 moles of H₂

Then, 1720 moles of N₂ will react with (1720 .3) /1 = 5160 moles of H₂

We have 5740 moles of hydrogen, so we have enough hydrogen. This is the excess reagent, so the limiting is the N₂

1 mol of N₂ can produce 2 moles of ammonia

Therefore 1720 moles of N₂ will produce (1720 . 2) /1 = 3440 moles of NH₃

We apply now, the Ideal Gases Law → P . V = n . R .T

V = (n . R . T) /P → V = (3440 mol . 0.082 L.atm/mol.K . 243K) / 1.38 atm

V = 49671 L

We confirm that the nitrogen was the limiting reactant

3 moles of H₂ need 1 mol of nitrogen to react

Therefore, 5740 moles of H₂ will react with (5740 . 1) /3 = 1913 moles of N₂

It was ok to say, that N₂ was the limiting reactant because we need 1913 moles in the reaction, and we only have 1720 moles

6 0

3 years ago

How many Mol are in 4.000 grams of MgCl2 ?

grigory [225]

Answer:

The number of mol is: 0, 042 mol in 4 grams of MgCl2

Explanation:

We calculate the weight of 1 mol of MgCl2:

Weight 1mol of MgCl2= weight Mg + (weight Cl)x 2=

24, 3 grams + 2 x 35, 5 grams = 95, 3 grams/mol MgCl2

95, 3 grams------1 mol MgCl2

4 grams -------x = (4 grams x1 mol MgCl2)/ 95, 3 grams= 0, 04197 mol MgCl2

4 0

3 years ago