Name the following molecule in the image

If a temperature increase from 10.0 ∘C to 22.0 ∘C doubles the rate constant for a reaction, what is the value of the activation

maria [59]

The activation energy barrier is 40.1 kJ·mol⁻¹

Use the Arrhenius equation

$\ln( \frac{k_2 }{k_1 }) = (\frac{E_{a} }{R })(\frac{ 1}{T_1} - \frac{1 }{T_2 })\\$

$\ln( \frac{2k }{k}) = (\frac{E_{a} }{\text{8.314 J} \cdot \text{K}^{-1} \text{mol}^{-1} })(\frac{ 1}{\text{283.15 K}} - \frac{1 }{\text{295.15 K }})\\$

$\ln2 = (\frac{ E_{a} }{\text{8.314 J} \cdot \text{K}^{-1} \text{mol}^{-1}}) \times 1.436 \times10^{-4}\\$

$\ln2 = E_{a} \times 1.727 \times 10^{-5} \text{ mol} \cdot \text{J}^{-1}$

$E_{a} = \frac{\ln2 }{ 1.727 \times10^{-5}\text{ mol} \cdot \text{J}^{-1}}\\$

$E_{a} = \text{40 100 J}\cdot\text{mol}^{-1} = \textbf{40.1 kJ}\cdot \textbf{mol}^{-1}$

3 0

3 years ago

When Rutherford ran his gold foil experiment, he expected to see results like those in the Plum Pudding Atom simulation. Instead

DENIUS [597]

Rutherford's result can not be explained on the basis of the plum pudding model because of the fact that, since some alpha particles were deflected, the atom must contain a small region with a strong electric charge.

The empirical study of the atom led to the emergence of several models of the atom. In the Plum - pudding model, the atom was regarded as a positively charged sphere with embedded negative charges.

This model can not interpret the Rutherford experiment since alpha particles were deflected. The deflection of alpha particles means that, the atom must contain a small region with a strong electric charge.

Learn more: brainly.com/question/730256

4 0

2 years ago

Can someone help me with that please?

True [87]

That is pretty tricky i don’t know

3 0

3 years ago

Investigating iron thiocyanate requires several chemicals which must be used with caution. Using knowledge of these chemicals, s

xxTIMURxx [149]

Answer:

1. False

2. False

3. True

4. False

5. False

6. False

Explanation:

1. Because HNO₃ is corrosive when handling it is important to use gloves and also protection glasses. If it touches the skin, then it's important to wash the area affected.

2. The dilute of acid is exothermic, so it releases a huge amount of heat. For precaution, the acid must be added at the water slowly stirring, so it can be controlled.

3. Fe(NO₃)₃ is an oxidizer and in contact with the skin it can cause irritation, so it must be handled carefully and with protection.

4. KSC can irritate the eyes, so it's toxic.

5. KSCN is not inflammable and it's not combustible, so when heated it will only change the state for gas, but it will not cause the evolution of cyanide gas.

6. If KSCN reacts with a strong base, it will dissociate, and the ions SCN⁻, which can't react with the OH⁻ ions of the base. So, it will only cause the evolution of cyanide gas if it reacts with a strong acid.

3 0

4 years ago

Calculate the molarity (M) of a solution containing 49.0 grams of H3PO4 in 500 mL of solution.

vlada-n [284]

Answer

The molarity (M) of the H3PO4 solution = 1.0 M

Explanation

Given:

Mass of H3PO4 = 49.0 grams

Volume of the solution = 500 mL = 500/1000 = 0.5 L

What to find:

The molarity (M) of the H3PO4 solution.

Step-by-step solution:

Step 1: Convert 49.0 grams H3PO4 to moles using the mole formula.

$Mole=\frac{Mass}{Molar\text{ }mass}$

The molar mass of H3PO4 = 97.994 g/mol

So,

$Mole=\frac{49.0\text{ }g}{97.994\text{ }g\text{/}mol}=0.50\text{ }mol$

Step 2: Calculate the molarity of the solution using the molarity formula.

$Molarity=\frac{Mole}{Volume\text{ }in\text{ }L}$

Putting mole = 0.50 mol and volume = 0.50L into the formula, we have;

$Molarity=\frac{0.50mol}{0.50L}=1.0\text{ }M$

The molarity (M) of the H3PO4 solution = 1.0 M

5 0

1 year ago