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Ad libitum [116K]
3 years ago
12

Mosses and liverworts are examples of ___________.

Chemistry
1 answer:
raketka [301]3 years ago
6 0
Seedless non-vascular plants!!
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How will the following changes affect the mole fraction of chlorine gas, χcl2, in the equilibrium mixture.?
sammy [17]
If the reaction is represented by:
PCl₃ + Cl₂ <-> PCl₅ (exothermic)

the mole fraction of chlorine in the equilibrium mixture will change according to the following:
Decrease the volume: decrease
Increase the temperature: increase
Increase the volume: increase
Decrease the temperature: decrease
4 0
3 years ago
What does C represent
zimovet [89]

Answer: the distance of energy from point d to e

8 0
3 years ago
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A 3.06 gram sample of an unknown hydrocarbon with empirical formula CH2O was found to contain 0.0170 moles of the substance. Wha
Yanka [14]

Answer:

180 amu

C₆H₁₂O₆

Explanation:

Step 1: Determine the molecular mass of the compound

The sample has a mass (m) of 3.06 g and it contains (n) 0.0170 moles. The molar mass M is:

M = m/n = 3.06/0.0170 mol = 180 g/mol

Then, the molecular mass is 180 amu.

Step 2: Determine the molar mass of the empirical formula.

M(CH₂O) = 1 × M(C) + 2 × M(H) + 1 × M(O)

M(CH₂O) = 1 × 12 g/mol + 2 × 1 g/mol + 1 × 16 g/mol = 30 g/mol

Step 3: Determine the molecular formula

First, we will determine "n" according to the following expression.

n = molar mass molecular formula / molar mass empirical formula

n = 180 g/mol / 30 g/mol = 6

The molecular formula is:

n × CH₂O = 6 × CH₂O = C₆H₁₂O₆

5 0
2 years ago
Convert 5 micrometers to meters.
k0ka [10]

Answer:

5e-6

Explanation:

7 0
3 years ago
Read 2 more answers
What is the molality of an aqueous solution containing FeCl3 (MM = 162.2 g/mol) with a mole fraction of FeCl3 of 0.15?
Natalka [10]

Answer:

10 m

Explanation:

The mole fraction of FeCl₃ of 0.15, that is, per mole of solution, there are 0.15 moles of FeCl₃ and 1 - 0.15 = 0.85 moles of water.

The molar mass of water is 18.02 g/mol. The mass corresponding to 0.85 moles is:

0.85 mol × 18.02 g/mol = 15 g = 0.015 kg

The molality of FeCl₃ is:

m = moles of solute / kilogram of solvent

m = 0.15 mol / 0.015 kg

m = 10 m

3 0
3 years ago
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