Which unit can be used to express the rate of a reaction?

Amphetamine (C9H13N) is a weak base with a pKb of 4.2.

QveST [7]

Weak base: [OH⁻] = √Kb.C

pKb = 4.2

$\tt Kb=10^{-4.2}$

c = concentration

MM Amphetamine (C9H13N) = 135.21 g/mol

c = 215 mg/L = (0.215 g : 135,21 g/mol) / L = 0.00159 mol/L = 1.59 x 10⁻³ mol/L

$\tt [OH^-]=\sqrt{10^{-4.2}\times 1.59\times 10^{-3}}=3.17\times 10^{-4}$

pOH = 4 - log 3.17

pH = 14 - (4 - log 3.17)

pH = 10 + log 3.17 = 10.50

4 0

2 years ago

How would an ash cloud from a volcano affect the flow of matter, cycling of energy, and organisms in an ecosystem?

Allisa [31]

Ash cloud from a volcano can greatly affect so many systems in the earth. The flow of matter is greatly affected by energy because without energy matter does not move. The cloud of valcano affects cycle of energy because it blocks the energy of the sun the enters the earth which will result to reducing temperature. Example of it the erruption of mt. pinatubo in ph, the earth temperature substanctailly reduces. The ecosystem will greatly be affected due to different temperature and different weather pattern that will occur that will be new to the orgnisms.

3 0

2 years ago

Watch the video to determine which of the following relationships correctly depict the relationship between pressure and volume

sergij07 [2.7K]

Answer : $P\propto \frac{1}{V}$ and $V\propto \frac{1}{P}$

Explanation:

According to Boyle's law, it states that pressure is directly proportional to the volume of the gas for a fixed amount of gas and at constant temperature.

The equation given by this law is:

$P_1V_1=P_2V_2$

where,

$P_1\text{ and }V_1$ are initial pressure and volume.

$P_2\text{ and }V_2$ are final pressure and volume.

Thus the correct relation for pressure and volume for a fixed amount of gas at constant temperature according to Boyle’s law is $P\propto \frac{1}{V}$ and $V\propto \frac{1}{P}$

7 0

3 years ago

What is the answer please and why is this answer

Ad libitum [116K]

The answer is c it was 1510 is 1950 and in 1990 it was 16000

5 0

3 years ago

Liquid octane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 63. g of octane is mi

Sever21 [200]

Answer:

52.1 g is the maximum mass of CO₂, that can be produced by this combustion

Explanation:

Mass of Octane: 63 g

Mass of O₂: 59.4 g

This is a combustion reaction where the products, are always water and CO₂. We define the equation:

2C₈H₁₈ (l) + 25O₂(g) → 16CO₂(g) + 18H₂O(g)

As we have, both mases of each reactant, we must define which is the limiting reagent. We convert the mass to moles:

63 g. 1mol / 114g = 0.552 moles

59.4 g . 1mol / 32g = 1.85 moles

Certainly, the limiting reagent is the oxygen:

2 moles of octane need 25 moles of O₂ to react

Therefore, 0.552 moles of octane must need (0.552 . 25) /2 = 6.9 moles of O₂ (I do not have enough moles of oxygen, I need 6.9 and I only got 1.85 moles)

When we know the limiting reagent we can do the calculations with the stoichiometry of the reaction:

25 moles of O₂ can produce 16 moles of CO₂

Therefore, 1.85 moles of O₂ may produce (1.85 . 16) /25 = 1.18 moles.

We convert the moles to mass to get the final answer:

1.18 mol . 44 g / 1mol = 52.1 g

5 0

2 years ago

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