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2 years ago

Which two factors must be equal when a chemical reaction reaches equilibrium?

1 answer:

3 0

The correct answer is the statement in option 3. Chemical equilibrium is achieved when both the reactants and the products are in concentrations that has no further tendency to change even as time goes to infinity. This happens when the forward reaction rate is the same as the reverse reaction rate.

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Help! I am in a hurry! Will get brainliest if correct!

stira [4]

Answer:

Should be 3

Iron, cobalt, and nickel

7 0

2 years ago

What is name of each shape

dimulka [17.4K]

Answer:

I am so sorry I do not know this

Explanation:

4 0

2 years ago

If 38.5 grams of potassium react with excess oxygen gas, how many grams of potassium oxide can be produced? 4K + O2 yields 2K2O

Lera25 [3.4K]

Answer:

46.40 g.

Explanation:

It is a stichiometric problem.

The balanced equation of the reaction: 4K + O₂ → 2K₂O.

It is clear that 4.0 moles of K reacts with 1.0 mole of oxygen produces 2.0 moles of K₂O.

We should convert the mass of K (38.5 g) into moles using the relation:

<em>n = mass / molar mass,</em>

n = (38.5 g) / (39.098 g/mol) = 0.985 mole.

<em>Using cross multiplication:</em>

4.0 moles of K produces → 2.0 moles of K₂O, from the stichiometry.

0.985 mole of K produces → ??? moles of K₂O.

∴ The number of moles of K₂O produced = (0.985 mole) (2.0 mole) / (4.0 mole) = 0.4925 mole ≅ 0.5 mole.

Now, we can get the mass of K₂O:

∴ mass = n x molar mass = (0.5 mole) (94.2 g/mol) = 46.40 g.

6 0

2 years ago

How many types of alkanol do we have

oksano4ka [1.4K]

I think you mean alcohol ? if so there are a few basic types including ethanol

6 0

3 years ago

15. What volume of CCI, (d = 1.6 g/cc) contain

anastassius [24]

Answer:

$\boxed{\text{(3) 9.6 L}}$

Explanation:

1. Moles of CCl₄

$n = 6.02 \times 10^{25} \text{ molecules} \times \dfrac{\text{1 mol}}{6.022 \times 10^{23}\text{ molecules}} = \text{100.0 mol}$

2. Molar mass of CCl₄

MM = 1 × 12.01 + 4 × 35.5 = 12.01 + 142 = 154.0 g/mol

3. Mass of CCl₄

$m =\text{100.0 mol} \times \dfrac{\text{154.0 g}}{\text{1 mol}} = \text{15 400 g}$

4. Volume of CCl₄

$V = \text{15 400 g} \times \dfrac{\text{1 cm}^{3}}{\text{1.6 g}} = \text{9600 cm}^{3}\\\\V = \text{9600 cm}^{3} \times \dfrac{\text{1 L}}{\text{1000 cm}^{3}} = \mathbf{{9.6 L}}\\\\\text{The volume of CCl$_{4}$ is } \boxed{\textbf{9.6 L}}$

4 0

2 years ago