Which of the following represents the most matter?

What are 3 examples of making diluted solutions in "regular" life?<br> what are examples of diluted?

katrin [286]

Answer:

Do it your self bro

Explanation:

You lazy bum do it yourself

6 0

3 years ago

In general, ionization energies increase across a period from left to right. Explain why the second ionization energy of Cr is h

rodikova [14]

Answer:So this leads to the fact that second ionization energy of chromium is higher as compared to that of Manganese because of the unavailability of electron in the outermost orbital in case of chromium so the second electron has to be removed form the stable half filled 3d orbital which requires more energy. Whereas in case of Manganese there is an electron available in outermost 4s orbital.

Explanation:

Ionization energy is the amount of energy that we require to remove an electron form an isolated gaseous atom.

As we move from left to right across a period electrons are added to the same outermost shell therefore the attraction between the electrons and nucleus increases since more number of negatively charged electron are attracted to the positively charged nucleus. This attraction leads to the decrease in atomic radii across a period and increase in ionization energy .

The increase in ionization energy occurs due to the fact that as the attraction between the nucleus and outermost electrons increases so the electrons are more tightly bound to the nucleus hence more amount of energy is required to ionize the electron which leads to increase in ionization energy.

The electronic configuration of Cr and Mn are:

Cr:[Ar]3d⁵4S¹

Mn:[Ar]3d⁵4S²

The electronic configuration of Cr and Mn after 1st ionization:

Cr:[Ar]3d⁵4S⁰

Mn:[Ar]3d⁵4S¹

The electronic configuration of Cr and Mn after 2nd ionization:

Cr:[Ar]3d⁴4S⁰

Mn:[Ar]3d⁵4S⁰

As we can see that that 3d orbital of Cr (Chromium) is half filled with 5 electrons in it and 4s orbital of Cr is also half-filled.

So when Cr is ionized for the first time then the electron from the half-filled 4s orbital will be removed .As the 1 electron present in outer most 4s orbital is removed so the 4s orbital now is completely vacant.

Now for the second ionization energy an electron ahs to be removed from half-filled 3d⁵ orbital. Hunds rule of maximum multiplicity states that the fully-filled or half-filled orbitals have maximum stability on account of symmetry and exchange energy.

So half-filled 3d⁵ orbital of Cr is very stable and hence to remove an electron from this would be require a lot of energy and hence the second ionization energy of chromium is higher than that of Manganese.

In case of Mn the 3d orbital is also half -filled as chromium but the 4s orbital contains two electrons. when we remove the first electron from this orbital then also there is 1 electron present in the 4s orbital . So for the second ionization of Mn the only electron left in 4s orbital will be removed as the removal of electron from a 4s orbital is much easier as it requires less amount of energy as compared to removal of a electron from stable half filled 3d orbital.

So this leads to the fact that second ionization energy of chromium is higher as compared to that of Manganese because of the unavailability of electron in the outermost orbital in case of chromium so the second electron has to be removed form the stable half filled 3d orbital which requires more energy. Whereas in case of Manganese there is an electron available in outermost 4s orbital.

3 0

3 years ago

For the reaction ? NO + ? O2 → ? NO2 , what is the maximum amount of NO2 which could be formed from 16.42 mol of NO and 14.47 mo

stira [4]

Answer:

1a. The balanced equation is given below:

2NO + O2 → 2NO2

The coefficients are 2, 1, 2

1b. 755.32g of NO2

2a. The balanced equation is given below:

2C6H6 + 15O2 → 12CO2 + 6H2O

The coefficients are 2, 15, 12, 6

2b. 126.25g of CO2

Explanation:

1a. Step 1:

Equation for the reaction. This is given below:

NO + O2 → NO2

1a. Step 2:

Balancing the equation. This is illustrated below:

NO + O2 → NO2

There are 2 atoms of O on the right side and 3 atoms on the left side. It can be balance by putting 2 in front of NO and 2 in front of NO2 as shown below:

2NO + O2 → 2NO2

The equation is balanced.

The coefficients are 2, 1, 2

1b. Step 1:

Determination of the limiting reactant. This is illustrated below:

2NO + O2 → 2NO2

From the balanced equation above, 2 moles of NO required 1 mole of O2.

Therefore, 16.42 moles of NO will require = 16.42/2 = 8.21 moles of O2.

From the calculations made above, there are leftover for O2 as 8.21 moles out of 14.47 moles reacted. Therefore, NO is the limiting reactant and O2 is the excess reactant.

1b. Step 2:

Determination of the maximum amount of NO2 produced. This is illustrated below:

2NO + O2 → 2NO2

From the balanced equation above, 2 moles of NO produced 2 moles of NO2.

Therefore, 16.42 moles of NO will also produce 16.42 moles of NO2.

1b. Step 3:

Conversion of 16.42 moles of NO2 to grams. This is illustrated below:

Molar Mass of NO2 = 14 + (2x16) = 14 + 32 = 46g/mol

Mole of NO2 = 16.42 moles

Mass of NO2 =?

Mass = number of mole x molar Mass

Mass of NO2 = 16.42 x 46

Mass of NO2 = 755.32g

Therefore, the maximum amount of NO2 produced is 755.32g

2a. Step 1:

The equation for the reaction.

C6H6 + O2 → CO2 + H2O

2a. Step 2:

Balancing the equation:

C6H6 + O2 → CO2 + H2O

There are 6 atoms of C on the left side and 1 atom on the right side. It can be balance by 6 in front of CO2 as shown below:

C6H6 + O2 → 6CO2 + H2O

There are 6 atoms of H on the left side and 2 atoms on the right. It can be balance by putting 3 in front of H2O as shown below:

C6H6 + O2 → 6CO2 + 3H2O

There are a total of 15 atoms of O on the right side and 2 atoms on the left. It can be balance by putting 15/2 in front of O2 as shown below:

C6H6 + 15/2O2 → 6CO2 + 3H2O

Multiply through by 2 to clear the fraction.

2C6H6 + 15O2 → 12CO2 + 6H2O

Now, the equation is balanced.

The coefficients are 2, 15, 12, 6

2b. Step 1:

Determination of the mass of C6H6 and O2 that reacted from the balanced equation. This is illustrated below:

2C6H6 + 15O2 → 12CO2 + 6H2O

Molar Mass of C6H6 = (12x6) + (6x1) = 72 + 6 = 78g/mol

Mass of C6H6 from the balanced equation = 2 x 78 = 156g

Molar Mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 15 x 32 = 480g

2b. Step 2:

Determination of the limiting reactant. This is illustrated below:

From the balanced equation above,

156g of C6H6 required 480g of O2.

Therefore, 37.3g of C6H6 will require = (37.3x480)/156 = 114.77g of O2.

From the calculations made above, there are leftover for O2 as 114.77g out of 126.1g reacted. Therefore, O2 is the excess reactant and C6H6 is the limiting reactant.

2b. Step 3:

Determination of mass of CO2 produced from the balanced equation. This is illustrated belowb

2C6H6 + 15O2 → 12CO2 + 6H2O

Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol

Mass of CO2 from the balanced equation = 12 x 44 = 528g

2b. Step 4:

Determination of the mass of CO2 produced by reacting 37.3g of C6H6 and 126.1g O2. This is illustrated below:

From the balanced equation above,

156g of C6H6 produced 528g of CO2.

Therefore, 37.3g of C6H6 will produce = (37.3x528)/156 = 126.25g of CO2

5 0

4 years ago

Chemical substances that dissolve in water or react with water to release ions are known as

disa [49]

The answer is electrolytes. Strong electrolytes like strong acids, strong bases and salts dissociate completely into ions when dissolved and no neutral molecules are present in their solution. Weak electrolytes like weak acids and weak bases do not completely ionize in water and some neutral molecules are present in their solution, while nonelectrolytes do not dissociate into ions when in solution at all.

8 0

4 years ago

How do electrostatic forces affect the electrons in an atom?

kkurt [141]

The answer is C.
Electrostatic force binds the the negatively charged to the positive charge atomic nucleus

3 0

2 years ago

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