<span>As the number of electrons added to the same principal energy level increases, atomic size generally A. increases.
</span>
<span>Because
as you add energy levels, the electrons are drawn to the nucleus of the atoms giving them a
lesser charge. Thus its atomic size also increases. </span>
Answer:
14. B 15. D 16. C 17. B
Explanation:
The spontaneous reaction that occurs when the cell operates is shown below:
⇒
We need to select the correct option from the list below for the following questions.
(A) Voltage increases. (B) Voltage decreases but remains > zero. (C) Voltage becomes zero and remains at zero. (D) No change in voltage occurs. (E) Direction of voltage change cannot be predicted without additional information.
14. A 50-milliliter sample of a 2-molar 
solution is added to the left beaker.
If a 50-milliliter sample of a 2-molar solution is added to the left beaker, the voltage decreases but its value remains greater than zero. The correct option is B
15. The silver electrode is made larger.
If the silver electrode is made larger, no change in the value of the voltage since we don't have the idea of the initial value. The correct option is D.
16. The salt bridge is replaced by a platinum wire.
If the salt bridge is replaced by a platinum wire, there will be no passage of electrons because electrons can't pass through a platinum wire. Therefore, the voltage will be zero and remains at zero. The correct option is C.
17. Current is allowed to flow for 5 minutes.
If current is allowed to flow for 5 minutes, the voltage decreases but its value remains greater than zero. The correct option is B.
<span>The best choice is hypochlorous acid nitrous acid (HNO2) because it has the nearest value of pK to the desired pH.
pKa of </span>nitrous acid<span> is 3.34
If we know pKa and pH values, we can calculate the required ratio of conjugate base (NO2⁻) to acid (HNO2) from the following equation:
pH=pKa + log(conc. of base)/( conc. of acid)
</span><span>3.19=3.34 + log c(NO2⁻)/c(HNO2)
</span><span>3.19 - 3.34 = log c(NO2⁻)/c(HNO2)
-0.15 = log c(NO2⁻)/c(HNO2)
c(NO2⁻)/c(HNO2) = 10⁰¹⁵ = 1.41
</span>
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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