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Montano1993 [528]
2 years ago
8

Why are the trends and exceptions to the trends in ionization energy observed? Check all that apply. Ionization energy tends to

increase down a group because the electrons get farther away from the nucleus. Ionization energy tends to increase across a period because the nuclear charge increases. Ionization energy tends to increase across a period because electrons are added to the same main energy level. The ionization energies of the elements in Group 16 tend to be slightly smaller than the elements in Group 15 because the fourth electron is added to an unfilled p orbital. The ionization energies of elements in Group 13 tend to be lower than the elements in Group 2 because the full s orbital shields the electron in the p orbital from the nucleus.
Chemistry
2 answers:
NNADVOKAT [17]2 years ago
5 0
The trends and exceptions to the trends in ionization energy observed includes;
B, ionization energy tends to increase across a period because the nuclear charge increases.
C, ionization energy tends to increase across a period because electrons are added to the same main energy level.
E, The ionization energies of elements in Group 13 tend to be lower than the elements in Group 2 because the full s orbital shields the electron, in the p orbital from the nucleus. 
Ionization energies measure the tendency of a neutral atom to resist the loss of electrons. It takes a considerable amount of energy, for example to remove an electron from a neutral fluorine atom to form a positively charged ion. <span />
Georgia [21]2 years ago
4 0

answers are

2) Ionization energy tends to increase across a period because the nuclear charge increases.

3)Ionization energy tends to increase across a period because electrons are added to the same main energy level.

5) The ionization energies of elements in Group 13 tend to be lower than the elements in Group 2 because the full s orbital shields the electron in the p orbital from the nucleus.


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Zinc and magnesium metal each reacts with hydrochloric acid to make chloride salts of the respective metals, and hydrogen gas. a
kirill115 [55]
M=11.20 g
m(H₂)=0.6854 g
M(H₂)=2.016 g/mol
M(Mg)=24.305 g/mol
M(Zn)=65.39 g/mol
w-?

m(Mg)=wm
m(Zn)=(1-w)m

Zn + 2HCl = ZnCl₂ + H₂
m₁(H₂)=M(H₂)m(Zn)/M(Zn)=M(H₂)(1-w)m/M(Zn)

Mg + 2HCl = MgCl₂ + H₂
m₂(H₂)=M(H₂)m(Mg)/M(Mg)=M(H₂)wm/M(Mg)

m(H₂)=m₁(H₂)+m₂(H₂)
m(H₂)=M(H₂)(1-w)m/M(Zn)+M(H₂)wm/M(Mg)=M(H₂)m{(1-w)/M(Zn)+w/M(Mg)}

m(H₂)=M(H₂)m{(1-w)/M(Zn)+w/M(Mg)}

(1-w)/M(Zn)+w/M(Mg)=m(H₂)/{M(H₂)m}

1/M(Zn)-w/M(Zn)+w/M(Mg)=m(H₂)/{M(H₂)m}

w(1/M(Mg)-1/M(Zn))=m(H₂)/{M(H₂)m}-1/M(Zn)

w=[m(H₂)/{M(H₂)m}-1/M(Zn)]/(1/M(Mg)-1/M(Zn))

w=0.583 (58.3%)
5 0
2 years ago
Find the enthalpy of neutralization of HCl and NaOH. 87 cm3 of 1.6 mol dm-3 hydrochloric acid was neutralized by 87 cm3 of 1.6 m
Vesna [10]

Answer : The correct option is, (A) -101.37 KJ

Explanation :

First we have to calculate the moles of HCl and NaOH.

\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=1.6mole/L\times 0.087L=0.1392mole

\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=1.6mole/L\times 0.087L=0.1392mole

The balanced chemical reaction will be,

HCl+NaOH\rightarrow NaCl+H_2O

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.1392 mole of HCl neutralizes by 0.1392 mole of NaOH

Thus, the number of neutralized moles = 0.1392 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = 87ml+87ml=174ml

\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 174ml=174g

Now we have to calculate the heat absorbed during the reaction.

q=m\times c\times (T_{final}-T_{initial})

where,

q = heat absorbed = ?

c = specific heat of water = 4.18J/g^oC

m = mass of water = 174 g

T_{final} = final temperature of water = 317.4 K

T_{initial} = initial temperature of metal = 298 K

Now put all the given values in the above formula, we get:

q=174g\times 4.18J/g^oC\times (317.4-298)K

q=14110.008J=14.11KJ

Thus, the heat released during the neutralization = -14.11 KJ

Now we have to calculate the enthalpy of neutralization.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy of neutralization = ?

q = heat released = -14.11 KJ

n = number of moles used in neutralization = 0.1392 mole

\Delta H=\frac{-14.11KJ}{0.1392mole}=-101.37KJ/mole

Therefore, the enthalpy of neutralization is, -101.37 KJ

3 0
2 years ago
Calculate the mass of MNO2 needed to produce 25.0 g of Cl2
Vedmedyk [2.9K]
I will solve this question assuming the reaction equation look like this:
<span>MnO2 + 4 HCl ---> MnCl2 + Cl2 + 2 H2O. 
</span>
For every one molecule of MnO2 used, there will be one molecule of Cl2 formed. If the molecular mass of MnO2 is 87g/mol and molecular mass of Cl2 is <span> 73.0 g/mol, the mass of MnO2 needed would be:
Cl mass/Cl molecular mass * MnO2 molecular mass=
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7 0
2 years ago
Read 2 more answers
How many valence electrons are in chlorodifluoromethane
Elis [28]
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7 0
2 years ago
A standard backpack is approximately 30cm x 30cm x 40cm. Suppose you find a hoard of pure gold while treasure hunting in the wil
Blizzard [7]

Explanation:

The dimensions of a standard backpack is 30cm x 30cm x 40cm

The mass of an average student is 70 kg

We know that, the density of gold is 19.3 g/cm³.

Let m be the mass of the backpack. So,

\text{density}=\dfrac{\text{mass}}{\text{volume}}\\\\m=d\times V\\\\m=19.3\ g/cm^3\times (30\times 30\times 40)\ cm^3\\\\m=694800\ g\\\\\text{or}\\\\m=694.8\ kg\approx 700\ kg

An average student has a mass of 70 kg. If we compare the mass of student and mass of backpack, we find that the backpack is 10 times of the mass of the student.

8 0
2 years ago
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