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dem82 [27]
2 years ago
13

Which of the following concerning second ionization energies is true?

Chemistry
1 answer:
Dennis_Churaev [7]2 years ago
5 0

Answer:

D

Explanation:

The fact is that the both elements belong to different groups in the periodic table. Mg is in group 12 while Al is in group 13. The outer most electron configuration for Mg2+=[Ne]3s0

Al+= [Ne]3s13p0

It is evident that in Al+, the second electron is to be removed from a filled 3s subshell which is energetically unfavourable. Unlike In Mg2+ where the two electrons are removed. There is always a tendency towards a quick loss of all the valence electrons in the valence shell. It will be more difficult to remove an electron from a filled shell.

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The Atomic Mass of Al is 26.98154 g/mol. Is it possible to have 5.0 × 10^-25 g of Al? Explain.
Lesechka [4]
<span>If a mole of aluminum weighs 26.98 grams, that means 1 atom of aluminum weighs = (26.98 g/mole) / (6.023 x 10^23 atoms/mole) = 4.479 x 10^-23 grams,

</span>so, it is not possible because 1 atom weighs that much we calculated which is <span>almost 100 times more than the amount you mentioned</span>

8 0
2 years ago
In the third period of the periodic table sodium is followed by magnesium aluminum silicon and phosphorus which of these element
wariber [46]
Answer:
            Phosphorous has the smallest atomic size.

Explanation:
                   As we know these elements belong to same period means there valence shell is the same. So moving from left to right along the period the shell number remains constant but the number of protons and electrons increases. So, due to increase in number of protons the nuclear charge increases hence attracts the valence electrons more effectively resulting in the decrease of atomic size.

Elements and their atomic radius are as follow,

<span><span>Magnesium          0.160 nm
</span><span>
Aluminium           0.130 nm
</span><span>
Silicon                  0.118 nm
</span><span>
Phosphorus         <span>0.110 nm</span></span></span>
6 0
2 years ago
A 14.3-cm3 sample of tin has a mass of 0.104 kg.
VladimirAG [237]
This is a true statement if it is density you are looking for... Density problem.....

Density is the ratio of the mass of an object to its volume.
D = m / V
D = 104g / 14.3 cm³ = 7.27 g/cm³ .............. to three significant digits

The conventions for the units of density is that grams per cubic centimeter (g/cm³) are usually used for solids, but will work for anything. Grams per milliliter (g/mL) are usually used for liquids and grams per liter (g/L) are for gases. Therefore, by convention, the units for tin (a solid) should be in grams per cubic centimeter.

Since 1 mL is equivalent to 1 cm³, then the density could be expressed as 7.27 g/mL.

The accepted value for the density of tin is 7.31 g/cm³
7 0
2 years ago
The density of o2 gas at 16 degrees Celsius and 1.27atm is?
velikii [3]

Answer:

The density of O₂ gas is 1.71 \frac{g}{L}

Explanation:

Density is a quantity that allows you to measure the amount of mass in a given volume of a substance. So density is defined as the quotient between the mass of a body and the volume it occupies:

density=\frac{mass}{volume}

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

So, you can get:

\frac{n}{V} =\frac{P}{R*T}

The relationship between number of moles and mass is:

n=\frac{mass}{molar mass}

Replacing:

\frac{\frac{mass}{molar mass} }{V} =\frac{P}{R*T}

\frac{mass}{V*Molar mass} =\frac{P}{R*T}

So:

\frac{mass}{V} =\frac{P*molar mass}{R*T}

Knowing that 1 mol of O has 16 g, the molar mass of O₂ gas is 32 \frac{g}{mol}.

Then:

\frac{mass}{V} =\frac{P*molar mass of O_{2} }{R*T}

In this case you know:

  • P=1.27 atm
  • molar mass of O₂= 32 \frac{g}{mol}.
  • R= 0.0821 \frac{atm*L}{mol*K}
  • T= 16 °C=  289 °K (0°C= 273°K)

Replacing:

density=\frac{mass}{V} =\frac{1.27atm*32\frac{g}{mol}  }{0.0821\frac{atm*L}{mol*K} *289 K}

Solving:

density= 1.71 \frac{g}{L}

<u><em>The density of O₂ gas is 1.71 </em></u>\frac{g}{L}<u><em></em></u>

3 0
2 years ago
A 2.25-g sample of magnesium nitrate, mg(no3)2, contains __________ mol of this compound.
MatroZZZ [7]
  Mg(No3)2  is  calculated   as  follows

moles  =  mass/molar  mass
the  molar  mass  of  Mg(NO3)2  is =  148 g/mol

moles  is  therefore= 2.25 g /  148  g/mol= 0.0152  moles

Mg(No3)2   contain  0.0152  moles  of  the   compound 
6 0
2 years ago
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