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Maslowich
3 years ago
6

Which action will slow down the rate of a reaction

Chemistry
1 answer:
Burka [1]3 years ago
4 0

Answer is - Decreasing surface area of a solid reactant, rate of reaction can be slowed down.

Explanation: The quantity of a reactant species consumed or the quantity of product species formed in unit time in a chemical reaction is called rate of reaction. With passage of time, concentration of reactant decreases and concentration of product increases. Rate of reaction depends on various factors like- concentration of reactant, temperature, nature of reactant, catalyst and surface area of reactant.

By increasing the concentration of the reactants, rate of the reaction increases.

Presence of catalyst increases the rate of reaction.

With increase in temperature, kinetic energy of molecules increases which results to more number of effective collisions, so, rate of reaction increases.

With increase in the surface area of reactant, more number of molecules can react at a time and by decreasing the surface area of solid reactant rate of the reaction will  diminish/ decrease.

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Molarity is measured in
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Suppose you were given a substance and asked to determine weather or not it was a plasma. Write the characteristics you would lo
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Electrical conductivity, electromagnetism, and temperature are the features that one would look for in order to determine plasma. Plasma refers to a hot ionized gas possessing high electrical conductivity. It is electrically neutral with negative and positive particles. It can be considered the most abundant form of matter in the universe.  

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3 years ago
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The H 2 produced in a chemical reaction is collected through water in a eudiometer. If the pressure in the eudiometer is 760.0 t
siniylev [52]

Answer:

the pressure (torr) of theH₂ gas is 736.2 torr

Explanation:

Given that

The H₂ gas  produced in a chemical reaction is collected through water in a eudiometer; during this process, the gas collected contains some droplets of water vapor along with these gas.

So; the total pressure in the eudiometer = Pressure in the H₂ gas  - Pressure of the water vapor

Where;

P_{Totsl} =  total pressure in the eudiometer = 760.0 torr

P_{H_2} = Pressure in the H₂ gas  = ???

P_{H_2O} = Pressure in the water vapor  = 23.8 torr

Now:

P_{Totsl} = P_{H_2} + P_{H_2O}

- P_{H_2} = + P_{H_2O} - P_{Totsl}

P_{H_2}  = - P_{H_2O} + P_{Totsl}

P_{H_2}  =  (- 23.8 + 760) torr

P_{H_2}  = 736.2 torr

Thus; the pressure (torr) of theH₂ gas is 736.2 torr

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4 years ago
. how do some deep-sea animals use their bioluminescence, or ability to give off light? (select all that apply.)
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3 years ago
Complete and balance the following equation: S(s)+HNO3(aq)→H2SO3(aq)+N2O(g)(acidic solution)
meriva

Answer:- The balanced equation is, 2HNO_3(aq)+2S(s)+H_2O(l)\rightarrow N_2O(g)+2H_2SO_3(aq) .

Solution:- Oxidation number of S is increasing from 0 to 4 and so it is oxidation. Oxidation number of N is decreasing from 5 to 1 and so it is reduction.

We write the oxidation and reduction half equations and balance them. The given reaction is taking place in an acidic medium.

First of all we balance all the atoms other than H and O. Then oxygen is balanced by adding H_2O and hydrogen is balanced by adding H^+ . Charge is balanced by adding electrons.

To makes the electrons equal for both the half equations we multiply the equation/equations by appropriate numbers.

Oxidation half equation:

S(s)\rightarrow H_2SO_3(aq)

S is already balanced. To balance O, we need to add three water molecules to the left side:

S(s)+3H_2O(l)\rightarrow H_2SO_3(aq)

For balancing hydrogen, we need to add 4 hydrogen ions to the right side:

S(s)+3H_2O(l)\rightarrow H_2SO_3(aq)+4H^+(aq)

Now to balance the charge we need to add 4 electrons to the right side:

S(s)+3H_2O(l)\rightarrow H_2SO_3(aq)+4H^+(aq)+4e^-

Reduction half equation:

HNO_3(aq)\rightarrow N_2O(g)

To balance N, we need to multiply left side by 2:

2HNO_3(aq)\rightarrow N_2O(g)

For balancing oxygen, we need to add 5 water molecules to the right side:

2HNO_3(aq)\rightarrow N_2O(g)+5H_2O(l)

To balance hydrogen, we need to add 8 hydrogen ions to the left side:

2HNO_3(aq)+8H^+(aq)\rightarrow N_2O(g)+5H_2O(l)

Now, for charge balance, we need to add 8 electrons to the left side:

2HNO_3(aq)+8H^+(aq)+8e^-\rightarrow N_2O(g)+5H_2O(l)

First half equation has 4 electrons and second half equation has 8 electrons.

To make the electrons equal, we need to multiply oxidation half equation by 2:

2S(s)+6H_2O(l)\rightarrow 2H_2SO_3(aq)+8H^+(aq)+8e^-

Now we add both of these two half equations and cancel out common species. What we get on doing this is:

2HNO_3(aq)+2S(s)+H_2O(l)\rightarrow N_2O(g)+2H_2SO_3(aq)



6 0
3 years ago
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