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3 years ago

PLEASE HELP ME!!

1 answer:

7 0

1, When temperature is increased the volume will also increase. this is because the particles will gain kinetic energy and bombard the walls of the container of the gas at a higher frequency, therefore, for the pressure to remain constant as per Charles' law, the volume will have to increase so that the rate of bombardment remains constant. This is explained by the Charles law which states that the volume of a gas is directly proportional to the absolute temperature provided pressure remains constant.

2. When temperature is Decreased the volume will also Decrease. this is because the particles will loose kinetic energy and bombard the walls of the container of the gas less frequently, therefore, for the pressure to remain constant as per Charles' law, the volume will have to reduce so that the rate of bombardment remains constant. This is explained by the Charles law which states that the volume of a gas is directly proportional to the absolute temperature provided pressure remains constant.

3. When temperature is increased the pressure will increase. This is because the gas particles gain kinetic energy and bombard the walls of the container more frequently. this is according to Pressure law which states that for a constant volume of a gas the pressure is directly proportional to absolute temperature

4. When temperature is decreased, pressure will decrease, This is because the gas particles lose kinetic energy and bombard the walls of the container less frequently. this is according to Pressure law which states that for a constant volume of a gas the pressure is directly proportional to absolute temperature

5. When particles are added, pressure will increase. This is because the bombardment per unit area also increases. Boyles law explains this, that at fixed temperature the volume of a gas is inversely proportional to the pressure.

6. When particles are removed, the pressure will decrease. This is because the bombardment per unit area also decreases. Boyle's law explains this, that at fixed temperature the volume of a gas is inversely proportional to the pressure.

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2 years ago

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3 years ago

The activation energy for proline isomerization of a peptide depends on the identity of the preceding residue and obeys Arrheniu

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Answer:

Activation energy of phenylalanine-proline peptide is 66 kJ/mol.

Explanation:

According to Arrhenius equation- $k=Ae^{\frac{-E_{a}}{RT}}$ , where k is rate constant, A is pre-exponential factor, $E_{a}$ is activation energy, R is gas constant and T is temperature in kelvin scale.

As A is identical for both peptide therefore-

$\frac{k_{ala-pro}}{k_{phe-pro}}=e^\frac{[E_{a}^{phe-pro}-E_{a}^{ala-pro}]}{RT}$

Here $\frac{k_{ala-pro}}{k_{phe-pro}}=\frac{0.05}{0.005}$ , T = 298 K , R = 8.314 J/(mol.K) and $E_{a}^{ala-pro}=60kJ/mol$

So, $\frac{0.05}{0.005}=e^{\frac{[E_{a}^{phe-pro}-(60000J/mol)]}{8.314J.mol^{-1}.K^{-1}\times 298K}}$

$\Rightarrow$ $E_{a}^{phe-pro}=65705J/mol=66kJ/mol$ (rounded off to two significant digit)

So, activation energy of phenylalanine-proline peptide is 66 kJ/mol

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3 years ago

Blackmail threat of informational disclosure is an example of which threat category?

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Answer:

Information Extortion.

Explanation:

Computer oriented crime also known as cyber crime that intentionally harm the victims. The complete data of the company or net banking information can be hacked easily by the cyber crime.

Different category of the threats are included in the cyber crime. The information extortion threat is the crime in which the hackers hack complete information and control the data of the victim. The criminal an black mail the victim regarding the hacked data.

Thus, the correct answer is information extortion.

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3 years ago

When 231. mg of a certain molecular compound X are dissolved in 65. g of benzene (CH), the freezing point of the solution is mea

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Answer:

Molar mass X = 18.2 g/mol

Explanation:

Step 1: Data given

Mass of compound X = 231 mg = 0.231 grams

Mass of benzene = 65.0 grams

The freezing point of the solution is measured to be 4.5 °C.

Freezing point of pure benzene = 5.5 °C

The freezing point constant for benzene is 5.12 °C/m

Step 2: Calculate molality

ΔT = i*Kf*m

⇒ΔT = the freezing point depression = 5.5 °C - 4.5 °C = 1.0 °C

⇒i = the Van't hoff factor = 1

⇒Kf = the freezing point depression constant for benzene = 5.12 °C/m

⇒m = the molality = moles X / mass benzene

m = 1.0 / 5.12 °C/m

m = 0.1953 molal

Step 3: Calculate moles X

Moles X = molality * mass benzene

Moles X = 0.1953 molal * 0.065 kg

Moles X = 0.0127 moles

Step 4: Calculate molar mass X

Molar mass X = mass / moles

Molar mass X = 0.231 grams / 0.0127 moles

Molar mass X = 18.2 g/mol

5 0

3 years ago