All categories

4 years ago

Which elements, when they have to, can have more than

Chemistry

1 answer:

inn [45]4 years ago

6 0

Elements in the third row can break the octet rule

You might be interested in

Next the students measured the density of aluminum metal. The accepted density for aluminum is 2.7 g/cm3. The students collected

katovenus [111]

Answer:

yo that's too long you should try and my it shorter

3 0

4 years ago

¿Que científico estableció que los electrones giran alrededor de un núcleo en varios niveles de energía?

svetlana [45]

Niels Bohr, hope it helps

6 0

3 years ago

600. ml of a gas at 50.0 kPa is compressed, at constant temperature, until its volume becomes 200. ml. What is the new pressure

Flura [38]

Answer:

<h3>The answer is 150 kPa</h3>

Explanation:

The new pressure can be found by using the formula for Boyle's law which is

$P_1V_1 = P_2V_2$

Since we are finding the new pressure

$P_2 = \frac{P_1V_1}{V_2} \\$

From the question we have

$P_2 = \frac{600 \times 50000}{200} = \frac{30000000}{200} \\ = 150000$

We have the final answer as

Hope this helps you

6 0

3 years ago

A student was assigned to take water samples from a lake his home . He measured the pH of one of the water samples to be 6.0 . W

Vinvika [58]

Answer: I'm guessing but I think this is the answer and also what I put in mine sorry if wrong

Explanation: I think it's "B.- Slightly Acidic" srry if it's too late-

8 0

3 years ago

The activation energy of an uncatalyzed reaction is 70 kJ/mol. When a catalyst is added, the activation energy (at 20 °C) is 42

denis23 [38]

Answer:

T = 215.33 °C

Explanation:

The activation energy is given by the Arrhenius equation:

$k = Ae^{\frac{-Ea}{RT}}$

<u>Where:</u>

k: is the rate constant

A: is the frequency factor

Ea: is the activation energy

R: is the gas constant = 8.314 J/(K*mol)

T: is the temperature

We have for the uncatalyzed reaction:

Ea₁ = 70 kJ/mol

And for the catalyzed reaction:

Ea₂ = 42 kJ/mol

T₂ = 20 °C = 293 K

The frequency factor A is constant and the initial concentrations are the same.

Since the rate of the uncatalyzed reaction (k₁) is equal to the rate of the catalyzed reaction (k₂), we have:

$k_{1} = k_{2}$

$Ae^{\frac{-Ea_{1}}{RT_{1}}} = Ae^{\frac{-Ea_{2}}{RT_{2}}}$ (1)

By solving equation (1) for T₁ we have:

$T_{1} = \frac{T_{2}*Ea_{1}}{Ea_{2}} = \frac{293 K*70 kJ/mol}{42 kJ/mol} = 488. 33 K = 215.33 ^\circ C$

Therefore, we need to heat the solution at 215.33 °C so that the rate of the uncatalyzed reaction is equal to the rate of the catalyzed reaction.

I hope it helps you!

4 0

4 years ago