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zaharov [31]
3 years ago
6

An increase in the surface area of reactants in a heterogeneous reaction will result in what?

Chemistry
1 answer:
marusya05 [52]3 years ago
4 0
It will result in an increase in the rate of rxn
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Element A has a half-life of 10 days. A scientist measures out 200 g of this substance. After 30 days has passed, the scientist
myrzilka [38]
N=N₀*2^(-t/T)

N₀=200 g
T=10 d
t=30 d

N=200*2^(-30/10)=25 g

25 g will remain
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An atom of oxygen has eight electrons and wight protons true or false
bazaltina [42]
I think the answer is true
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What is equal to the number of particles in 24 grams of carbon-12? 2 moles 2:1 mole ratio 2:1 molar mass 1:2 molar volume
mariarad [96]

Answer:

The answer to your question is 2 moles of Carbon

Explanation:

Carbon-12 means it has a molecular mass of 12g / mol. Mol is defined as the number of grams of an atom in one mol.

To solve this problem, we use proportions

                                    12 g of Carbon ---------------- 1 mol of Carbon

                                     24 g of Carbon --------------   x

                                     

                                      x = ( 24 g x 1 mol ) / 12 g

                                      x = 24 / 12

                                     x = 2 mol of Carbon

8 0
3 years ago
When 2 grams of powdered lead (IV) oxide was added to 100 cm3 of hydrogen peroxide, water and oxygen were produced. Lead (IV) ox
kirza4 [7]

Answer:

The right answer is:

Replacing the powdered lead oxide with its large crystals

Removing lead (IV) oxide from the reaction mixture

Using 1.0 gram of lead (IV) oxide

Explanation:

Based on the given information this reaction is the catalytic decomposition of H₂O₂ into water and oxygen  using Lead (IV) oxide as a catalyst.

  1. The catalyst surface area is directly proportional to the reaction rate
  • So, Replacing the powdered lead oxide with its large crystals would decrease the reaction rate due to the has larger surface area than its large crystals.

     2. Also, Removing lead (IV) oxide from the reaction mixture  the reaction rate decreased because as the catalyst is removed.

     3.  Using 50 cm³ of hydrogen peroxide  doesn't affect the rate because the concentration of the reactant doesn't change.

     4. Using 1.0 gram of lead (IV) oxide would decrease the reaction rate because the amount of catalyst decreased

So, The right answer is:

Replacing the powdered lead oxide with its large crystals

Removing lead (IV) oxide from the reaction mixture

Using 1.0 gram of lead (IV) oxide

7 0
4 years ago
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The frames of reference can be used for the same motion by —————————————————————————————ERROR——————————————
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