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4 years ago

Addition of a catalyst can speed up a reaction by providing an alternate reaction pathway that has a

Chemistry

1 answer:

kkurt [141]4 years ago

5 0

Answer:

Addition of a catalyst can speed up a reaction by providing an alternate reaction pathway that has a lower activation energy

Explanation:

A catalyst is an agent that increases the rate of a chemical reaction by providing an alternate pathway for the reaction that requires a lower activation energy. As the requirement for activation energy is less in the presence of a catalyst, there are more reactant particles becoming involved in the chemical reaction and as such there are more products formed per unit time, or there is an increase in the rate of the reaction

Example of catalyst include

1. Addition of potassium permanganate to hydrogen peroxide to aid in the rapid decomposition into water and oxygen

2. Platinum serves as a catalyst in the complete combustion of carbon monoxide into carbon dioxide.

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How many electrons are in p orbitals in Si (Silicon)?

insens350 [35]

Answer:

6 electrons

Explanation:

The p orbital can hold up to six electrons. We'll put six in the 2p orbital and then put the next two electrons in the 3s.

8 0

3 years ago

Aclosed system contains an equimolar mixture of n-pentane and isopentane. Suppose the system is initially all liquid at 120°C an

garri49 [273]

Explanation:

The given data is as follows.

T = $120^{o}C$ = (120 + 273.15)K = 393.15 K,

As it is given that it is an equimolar mixture of n-pentane and isopentane.

So, $x_{1}$ = 0.5 and $x_{2}$ = 0.5

According to the Antoine data, vapor pressure of two components at 393.15 K is as follows.

$p^{sat}_{1}$ (393.15 K) = 9.2 bar

$p^{sat}_{1}$ (393.15 K) = 10.5 bar

Hence, we will calculate the partial pressure of each component as follows.

$p_{1} = x_{1} \times p^{sat}_{1}$

= $0.5 \times 9.2 bar$

= 4.6 bar

and, $p_{2} = x_{2} \times p^{sat}_{2}$

= $0.5 \times 10.5 bar$

= 5.25 bar

Therefore, the bubble pressure will be as follows.

P = $p_{1} + p_{2}$

= 4.6 bar + 5.25 bar

= 9.85 bar

Now, we will calculate the vapor composition as follows.

$y_{1} = \frac{p_{1}}{p}$

= $\frac{4.6}{9.85}$

= 0.467

and, $y_{2} = \frac{p_{2}}{p}$

= $\frac{5.25}{9.85}$

= 0.527

Calculate the dew point as follows.

$y_{1}$ = 0.5, $y_{2}$ = 0.5

$\frac{1}{P} = \sum \frac{y_{1}}{p^{sat}_{1}}$

$\frac{1}{P} = \frac{0.5}{9.2} + \frac{0.5}{10.2}$

$\frac{1}{P}$ = 0.101966 $bar^{-1}$

P = 9.807

Composition of the liquid phase is $x_{i}$ and its formula is as follows.

$x_{i} = \frac{y_{i} \times P}{p^{sat}_{1}}$

= $\frac{0.5 \times 9.807}{9.2}$

= 0.5329

$x_{z} = \frac{y_{i} \times P}{p^{sat}_{1}}$

= $\frac{0.5 \times 9.807}{10.5}$

= 0.467

4 0

4 years ago

What 3 pieces of information do you need when describing an object's motion?

Aleksandr-060686 [28]

Answer: To describe an object’s motion, you must have a distance, km, mm, m, a speed, and time to show how long, seconds, minuets, hours, days.

7 0

3 years ago

You have a radioactive isotope with a mass of 400 grams. The sample has a half life of 5 days. How much is left after 25 days ha

Alexeev081 [22]

Answer:

Amount left after 25 days = 12.5 g

Explanation:

Given data:

Mass of sample = 400 g

Half life of sample = 5 days

Mass left after 25 days = ?

Solution:

First of all we will calculate the number of half lives passes in given time period.

Number of half lives = Time elapsed / Half life

Number of half lives = 25 days/ 5 days

Number of half lives = 5

At time zero = 400 g

At 1st half life = 400 g/2 = 200 g

At 2nd half life = 200 g/2 = 100 g

At 3rd half life = 100 g/2 = 50 g

At 4th half life = 50 g/2 = 25 g

At 5th half life = 25 g/2 = 12.5 g

4 0

3 years ago

Help please chemistry will give brainliest

Jlenok [28]

Answer:

Explanation:

The Ideal Gas Law states that PV=nRT.

Rearrange that into P/n=RT/V.

In this case, the cylinder is rigid so the volume, V, does not change.

Temperature does not change either.

Out of 450 grams of gas, 150 grams leak out. So only 450-150 = 300 grams is left.

n is number of moles which is dependent on mass:

n1/n2 = 450/300 = 3/2

P1/n1 = RT/V = P2/n2

P2 = P1/n1*n2

= 7.2/3*2

= 4.8 atmosphere

6 0

3 years ago

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