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Sloan [31]
2 years ago
7

It takes energy to ionize any atom. It takes progressively more and more energy for each successive electron that is removed and

the increase is somewhat smooth. For the removal of which electron does boron exhibit a sudden marked increase in its ionization energy?
Chemistry
1 answer:
IrinaVladis [17]2 years ago
5 0

Answer:

Removal of Third Electron

Explanation:

a major jump is required to remove the third electron. In general, successive ionization energies always increase because each subsequent electron is being pulled away from an increasingly more positive ion.

Ionization energy increases from bottom to top within a group, and increases from left to right within a period.

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Write the expression for the equilibrium constant Kp for the following reaction.Enclose pressures in parentheses and do NOT writ
maxonik [38]

<u>Answer:</u> The expression for K_p is written below.

<u>Explanation:</u>

Equilibrium constant in terms of partial pressure is defined as the ratio of partial pressures of the products and the reactants each raised to the power their stoichiometric ratios. It is expressed as K_p

For a general chemical reaction:

aA+bB\rightleftharpoons cC+dD

The expression for K_p is written as:

K_p=\frac{P_{C}^c\times P_{D}^d}{P_{A}^a\times P_{B}^b}

The partial pressure for solids and liquids are taken as 1.

For the given chemical equation:

NH_4HS(s)\rightleftharpoons NH_3(g)+H_2S(g)

The expression for K_p for the following equation is:

K_p=\frac{(P NH_3)\times (P H_2S)}{(P NH_4HS)}

The partial pressure of NH_4HS will be 1 because it is solid.

So, the expression for K_p now becomes:

K_p=\frac{(P NH_3)\times (P H_2S)}{1}

Hence, the expression for K_p is written above.

5 0
2 years ago
William adds two values, following the rules for using significant figures in computations. He should write the sum of these two
Lunna [17]
<span>when it comes to adding or subtracting numbers, his final answer should have the same number of decimal places as the least precise value.
For example if you add 2 numbers; 10.443 + 3.5 , 10.443 has 3 decimal places and 3.5 has only one decimal place.
Therefore 3.5 is the less precise value.
So when adding these 2 values the final answer should have only one decimal place.
after adding we get 13.943 but it can have upto one decimal place. then the second decimal place is less than 5 so the answer should be rounded off to 13.9.
the answer is the same number of decimal places as the least precise value</span>
6 0
2 years ago
Read 2 more answers
Discuss some of the biotic (living) and abiotic (nonliving) factors in the chimps’ ecosystem that affect their behavior.
Genrish500 [490]

Answer:

Diet:  fruit, leaves, bark, stems, seeds, eggs, insects, birds, small to medium sized primates - red tail monkeys, yellow baboons, bushbuck and warthogs.

Environmental Relationship - The chimpanzee keeps the plants it eats short, moves dirt around which helps things living in the dirt, keeps bird and small monkey populations that it eats from overpopulating.

Different biotic and abiotic factors affect why the chimps live where they do. (Spatial Relationships)

Explanation:

8 0
2 years ago
Calculate the heat of reaction, ΔH°rxn, for overall reaction for the production of methane, CH4.
Lesechka [4]

<u>Answer:</u> The enthalpy of the reaction for the production of CH_4 is coming out to be -74.9 kJ

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as \Delta H^o

The equation used to calculate enthalpy change is of a reaction is:  

\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]

For the given chemical reaction:

C(s)+2H_2(g)\rightarrow CH_4(g)

The equation for the enthalpy change of the above reaction is:

\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(CH_4(g))})]-[(1\times \Delta H^o_f_{(C(s))})+(2\times \Delta H^o_f_{(H_2(g))})]

We are given:

\Delta H^o_f_{(C(s))}=0kJ/mol\\\Delta H^o_f_{(H_2)}=0kJ/mol\\\Delta H^o_f_{CH_4}=-74.9kJ/mol

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(1\times (-74.9))]-[1\times 0)+(2\times 0)]\\\\\Delta H^o_{rxn}=-74.9kJ

Hence, the enthalpy of the reaction for the production of CH_4 is coming out to be -74.9 kJ

3 0
2 years ago
Calcium-45 has a half-life of 162.7 days. If four half-lives have elapsed, how much time has passed?
vovikov84 [41]
Half-life<span> is the time required for a quantity to reduce to half its initial value. </span><span>If four half-lives have elapsed for calcium-45, then it would be 4x162.7 = 650.8 days have passed. Hope this answers the question. Have a nice day. Feel free to ask more questions.</span>
7 0
2 years ago
Read 2 more answers
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