A first-order reaction is 81omplete in 264s.The half-life for this reaction (i) t 1/2 = =3.465×10 −3 s.to reach 95% Completion = 285 s.
To measure reaction rates, chemists initiate the reaction, measure the concentration of the reactant or product at different times as the reaction progresses,
For a 0-order response, the mathematical expression that may be employed to determine the half of life is: t1/2 = [R]0/2k. For a first-order reaction, the half of-existence is given by: t1/2 = zero.693/ok. For a 2d-order response, the method for the half-life of the response is: 1/okay[R]0
The 1/2-life of a response (t1/2), is the quantity of time needed for a reactant concentration to lower via half of compared to its initial awareness. Its software is used in chemistry and medicine to are expecting the awareness of a substance over time
Half of the lifestyles is the time required for exactly 1/2 of the entities to decay 50%.
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Atomic size decreases in a period but the ionization energy and electronegativity increases across a period.
<h3>
Describe the trends in the atomic size, ionization energy and electronegativity?</h3>
Atomic radius decreases across a period because of nuclear charge increases whereas atomic radius of atoms generally increases from top to bottom within a group because there is again an increase in the positive nuclear charge.
Ionization energy increases when we move from left to right across an period and decreases from top to bottom.
Electronegativity also increases from left to right across a period and decreases from top to bottom.
So we can conclude that atomic size decreases in a period but the ionization energy and electronegativity increases across a period.
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The value of ΔG° at this temperature is -18034.18 J/mol
Calculation,
Given information
formation constant (Kf)= 1.7 × 
Universal gas constant (R) = 8.314 J/K• mol
Temperature = 25° C = 25 °C + 273 = 300 K
Formula used:
ΔG° = -RT㏑Kf
By putting the valur of R,T, Kf we get the value of ΔG°
ΔG° = - 8.314 J/K• mol×300K㏑ 1.7 ×
ΔG° = -2494.2㏑ 1.7 × = -18034.18 J/mol
So, change in standard Gibbs's free energy is -18034.18 J/mol
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Following laboratory safety protocols such as wearing personal protective equipment will protect John when the accident occurred.
<h3>What are laboratory safety protocols?</h3>
Laboratory safety protocols are the protocols put in place to ensure safety in the laboratory.
Laboratory safety protocols include the following:
- always wear personal protective equipment in the laboratory
- do not play in the laboratory
- do not eat in the laboratory
Following laboratory safety protocols will help protect us from accidents which occur in the laboratory.
What happened when john was carefully pouring a chemical into a beaker when the beaker slips and breaks is an example of laboratory accident.
Wearing personal protective equipment will protect John.
In conclusion, following laboratory safety protocols will protect us when accidents occur in the laboratory.
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Note that the complete question is given as follows:
John is carefully pouring a chemical into a beaker when the beaker slips and breaks. How would laboratory safety protocols help John?
