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Hoochie [10]
1 year ago
8

the rate of a reaction increases when a catalyst has been added to the reaction mixture. the catalyst a) increases the kinetic e

nergy of the reactants. b) decreases the rate of the reverse reaction. c) increases the number of collisions between reactants. d) alters the enthalpy of reaction. e) provides a new mechanism for the reaction to proceed by.
Chemistry
1 answer:
Karolina [17]1 year ago
8 0

The rate of  the reaction increases when a catalyst has been added to the reaction mixture by e) providing a new mechanism for the reaction to procced by

Catalyst, in chemistry, a substance that speeds up a reaction without itself being consumed. Enzymes are natural catalysts responsible for many important biochemical reactions.

Most solid catalysts are oxides, sulfides, halides of metals or metallic elements, and metalloid elements boron, aluminum and silicon. Gas and liquid catalysts are usually used in pure form or in combination with suitable carriers or solvents. Solid catalysts are usually dispersed in another material known as a catalyst support.

Learn more about the catalyst in

brainly.com/question/28813725

#SPJ4

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3 years ago
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4 0
2 years ago
15.89 percent carbon, 21.18 percent oxygen, and 62.93 percent osmium. what is the empirical formula
otez555 [7]

Answer:

OsCO or COOs

Explanation:

Data given

Carbon = 15.89 %

Oxygen = 21.18 %

Osmium = 62.93%

Empirical formula = ?

Solution:

First find the masses of each component

Consider total compound is 100g

As we now

mass of element = % of component

So,

15.89 g of C     = 15.89 % Carbon

21.18 g of O      =   21.18 % Oxygen

62.93 g of Os  =   62.93% Osmium

Now convert the masses to moles

For Carbon

Molar mass of C = 12 g/mol

                  no. of mole = mass in g / molar mass

Put value in above formula

                  no. of mole =  15.89 g/ 12 g/mol

                  no. of mole =  1.3242

mole of C = 1.3242

For Oxygen

Molar mass of O = 16 g/mol

                  no. of mole = mass in g / molar mass

Put value in above formula

                  no. of mole =  21.18 g/ 16 g/mol

                  no. of mole =  

mole of O = 1.3238

For Os

Molar mass of Os = 190 g/mol

                  no. of mole = mass in g / molar mass

Put value in above formula

                  no. of mole =  62.93 g/ 190 g/mol

                  no. of mole =  

mole of Os = 1.3312

Now we have values in moles as below

C = 1.3242

O = 1.3238

Os = 1.3312

Divide the all values on the smalest values to get whole number ratio

C = 1.3242 /1.3238 = 1.0003

O = 1.3238 /1.3238 = 1

Os = 1.3312 /1.3238 = 1.0056

So all have round value 1 mole

C = 1

O = 1

Os = 1

So the empirical formula will be (OsCO) i.e. all 3 atoms in simplest small ratio

7 0
3 years ago
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