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3 years ago

How does increasing temperature increase the number of reactions?

Chemistry

1 answer:

antiseptic1488 [7]3 years ago

7 0

Answer:

Increasing the temperature increases reaction rates because of the disproportionately large increase in the number of high energy collisions. It is only these collisions (possessing at least the activation energy for the reaction) which result in a reaction

Explanation:

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Which of the following correctly describes the general process of coal mining and refining? Select one: a. Coal is mined, washed

e-lub [12.9K]

The general process for coal mining and refining is option C.

Coal mining is carried out by two main processes based on the depth, quality, thickness, and structure of coal seam. Based on the depth, coal seam of less than 180ft to the surface is surface mined while coal seems between 180ft.-300ft. are deep mined. The coal mined is transported to a nearby processing plant using a conveyor, train, ship, etc. where it is washed to remove other component like dirt, sulphur, rocks, etc. and afterwards dried.

4 0

4 years ago

The average molecular weight for element X is 59.97 g/mol. There are two known isotopes of element X, one weighing 59 g/mol, and

Temka [501]

Answer:

The correct answer is:48.5% X-59, 51.5% X-61

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

Let the fractional abundance of X-59 isotope be 'x'. So, fractional abundance of X-61 isotope will be '1 - x'

For X-59 isotope:

Mass of X-59 isotope =59 g/mol

Fractional abundance of X-59 isotope = x

For X-61 isotope:

Mass of X-61 isotope = 61 g/mol

Fractional abundance of X-61 isotope = 1 - x

Average atomic mass of chlorine = 35.4527 amu

Putting values in equation 1, we get:

$59.97 g/mol=[(59 g/mol\times x)+(61 g/mol\times (1-x))]\\\\x=0.515$

Percentage abundance of X-59 isotope = $0.515\times 100=51.5\%$

Percentage abundance of X-61 isotope = $(1-0.515)=0.485\times 100=48.5\%$

Hence, the percentage abundance of both the isotopes X-59and X-61 are 51.5% and 48.5% respectively.

4 0

3 years ago

Help please! I’ll mark brainliest to who ever answers this first!!

Nesterboy [21]

Answer:

paleontologists is the answer to scientist who studies fossil

6 0

3 years ago

How many grams of dry nh4cl need to be added to 2.00 l of a 0.600 m solution of ammonia, nh3, to prepare a buffer solution that

Brrunno [24]

Nh4cl = 117.165 gram ( solution is attached )

8 0

4 years ago

There are several aromatic compounds with the formula c8h9cl. Draw those that have a monosubstituted ring.

anygoal [31]

Answer:

As shown in the attachment.

Explanation:

In this case, there are two isomers possible. Isomers are compounds that have the same molecular formular but different structural formular.

Considering the hydrocarbon C8H10, which has 8C expected saturated hydrocarbon. however when there is an excess of 8H, we have C8H18 which is the formular of octane, a member of the alkane family and from monosubstitution reaction which involves reaction with halogens (particularly Chlorine), the reaction will be a repeated substitution reaction where chlorine will be replaced with each of the hydrogen.

8 0

4 years ago