Which statement is true of the carbon atoms that make up a diamond

One question I still have is regarding the metric system. I don't have a clear understanding of what valence electrons are. Do v

Allushta [10]

Valence electrons is an outer shell electron that is associated with an atom, and that can participate in the formation of a chemical bond if the outer shell is not closed. In a single covalent bond, both atoms in the bond contribute one valence electron in order to form a shared pair.

The presence of valence electrons can determine the elements chemical properties, such as its valence—whether it may bond with other elements and, if so, how readily and with how many. In this way, a given element's reactivity is highly dependent upon its electronic configuration. For a main group element, a valence electron can exist only in the outermost electron shell; in a transition metal, a valence electron can also be in an inner shell.
An atom with a closed shell of valence electrons (corresponding to an electron configuration s2p6 for main group elements) tends to be chemically inert. Atoms with one or two valence electrons more than a closed shell are highly reactive due to the relatively low energy to remove the extra valence electrons to form a positive ion. An atom with one or two electrons less than a closed shell is reactive due to its tendency either to gain the missing valence electrons and form a negative ion, or else to share valence electrons and form a covalent bond.

Similar to a core electron, a valence electron has the ability to absorb or release energy in the form of a photon. An energy gain can trigger the electron to move (jump) to an outer shell; this is known as atomic excitation. Or the electron can even break free from its associated atom's shell; this is ionization to form a positive ion. When an electron loses energy (thereby causing a photon to be emitted), then it can move to an inner shell which is not fully occupied.

When forming ions, elements typically gain or lose the minimum number of electrons necessary to achieve a full octet. For example, fluorine has seven valence electrons, so it is most likely to gain one electron to form an ion with a 1- charge.

3 0

3 years ago

Which determines the reactivity of an alkali metal?

adelina 88 [10]

Answer:

its ability to lose electron

3 0

3 years ago

Read 2 more answers

1) Aluminum sulphate can be made by the following reaction: 2AlCl3(aq) + 3H2SO4(aq) Al2(SO4)3(aq) + 6 HCl(aq) It is quite solubl

kolezko [41]

Answer:

88.9%

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

2AlCl3(aq) + 3H2SO4(aq) —> Al2(SO4)3(aq) + 6HCl(aq)

Step 2:

Determination of the masses of AlCl3 and H2SO4 that reacted and the mass of Al2(SO4)3 produced from the balanced equation.

Molar mass of AlCl3 = 27 + (35.5x3) = 133.5g/mol

Mass of AlCl3 from the balanced equation = 2 x 133.5 = 267g

Molar mass of H2SO4 = (2x1) + 32 + (16x4) = 98g/mol

Mass of H2SO4 from the balanced equation = 3 x 98 = 294g

Molar mass of Al2(SO4)3 = (27x2) + 3[32 + (16x4)]

= 54 + 3[32 + 64]

= 54 + 3[96] = 342g/mol

Mass of Al2(SO4)3 from the balanced equation = 1 x 342 = 342g

Summary:

From the balanced equation above,

267g of AlCl3 reacted with 294g of H2SO4 to produce 342g of Al2(SO4)3.

Step 3:

Determination of the limiting reactant. This is illustrated below:

From the balanced equation above,

267g of AlCl3 reacted with 294g of H2SO4.

Therefore, 25g of AlCl3 will react with = (25 x 294)/267 = 27.53g of H2SO4.

From the calculations made above, we see that only 27.53g out 30g of H2SO4 given were needed to react completely with 25g of AlCl3.

Therefore, AlCl3 is the limiting reactant and H2SO4 is the excess.

Step 4:

Determination of the theoretical yield of Al2(SO4)3.

In this case we shall be using the limiting reactant because it will produce the maximum yield of Al2(SO4)3 since all of it is used up in the reaction.

The limiting reactant is AlCl3 and the theoretical yield of Al2(SO4)3 can be obtained as follow:

From the balanced equation above,

267g of AlCl3 reacted to produce 342g of Al2(SO4)3.

Therefore, 25g of AlCl3 will react to produce = (25 x 342) /267 = 32.02g of Al2(SO4)3.

Therefore, the theoretical yield of Al2(SO4)3 is 32.02g

Step 5:

Determination of the percentage yield of Al2(SO4)3.

This can be obtained as follow:

Actual yield of Al2(SO4)3 = 28.46g

Theoretical yield of Al2(SO4)3 = 32.02g

Percentage yield of Al2(SO4)3 =..?

Percentage yield = Actual yield /Theoretical yield x 100

Percentage yield = 28.46/32.02 x 100

Percentage yield = 88.9%

Therefore, the percentage yield of Al2(SO4)3 is 88.9%

3 0

4 years ago

Hydrogen is manufactured on an industrial scale by this sequence of reactions: CH2 (g) + H2O(g)=CO (g) + 3H2 (g) K1 CO (g) + H2O

Flauer [41]

Answer:

K = K1×K2 = [CO2] [H2]⁴ / [H₂O]² [CH4]

Explanation:

Based on the reactions:

CH2 (g) + H2O(g) ⇄ CO (g) + 3H2 (g) K1

CO (g) + H2O (g) ⇄ CO2 (g)+H2(g) K2

The sum of both reactions is:

CH4 (g)+2H2O (g) ⇄ CO2(g)+4H2(g) And K of the reaction is: K = K1×K2

K is defined as the ratio between concentrations of products and reactans. Each compound must be elevated to its coefficient in the reaction. That is:

4 0

4 years ago

I need help on 18b. i’m confused on how you would work it.

AfilCa [17]

For a and b, you need to divide it by Avogadro’s number to find the answer.

a. (6.022x10^23)/6.022x10^23 = 1 mole of Ne
b. (3.011x10^23)/6.022x10^23 = 0.5 moles of Mg

For c and d, you’ll use the mass provided divided by the molar mass to find the number of moles.
Pb molar mass = 207.2 g/mol

c. (3.25x10^5)/207.2 = 1.57x10^3 moles of Pb
For d, I can’t tell if is Cu, C or something else but you can follow the steps above to solve the problem.

4 0

3 years ago