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3 years ago

Alkali metals, alkaline earth metals, and aluminum all form ions with positive charges equal to the

Chemistry

2 answers:

alisha [4.7K]3 years ago

7 0

Answer:

b. group number.

Explanation:

The alkali metals are located in group 1 of the periodic table and are the following: lithium, sodium, potassium, rubidium, cesium and francium and form ions with a positive charge because they are in group 1 of the periodic table .

The alkaline earth metals are located in group 2 of the periodic table and are the following: beryllium (Be), magnesium (Mg), calcium (Ca), strontium (Sr), barium (Ba) and radius (Ra) and form ions with two positive charges because they are in group 2 of the periodic table.

The aluminum is embedded within the set of other metals and belongs to group 3A and therefore can form an ion with 3 positive charges.

Amiraneli [1.4K]3 years ago

6 0

B. group number is the correct answer

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2 years ago

As noted in Assignment 1.4.1., alkylation of benzene with 1-chlorobutane in the presence of AlCl3 gives both butylbenzene and (1

Sunny_sXe [5.5K]

Answer:

Use the Bromotriflouride catalyst, BF₃

Explanation:

The BF₃ is most likely to yield less desired side products. The effect lies in the reaction mechanism.

BF₃ is a Lewis acid. Its role is to promote the ionization of the HF. This is achieved through the electrophilic mechanism. The reaction mechanism is as follows:

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3 years ago

Where the oxygen comes from the air (21% O2 and 79% N2). If oxygen is fed from air in excess of the stoichiometric amount requir

guajiro [1.7K]

Answer:

$y_{O2} =4.3$ %

Explanation:

The ethanol combustion reaction is:

$C_{2}H_{5} OH+3O_{2}$ → $2CO_{2}+3H_{2}O$

If we had the amount (x moles) of ethanol, we would calculate the oxygen moles required:

$x*1.10(excess)*\frac{3 O_{2}moles }{etOHmole}$

Dividing the previous equation by x:

$1.10(excess)*\frac{3 O_{2}moles}{etOHmole}=3.30\frac{O_{2}moles}{etOHmole}$

We would need 3.30 oxygen moles per ethanol mole.

Then we apply the composition relation between O2 and N2 in the feed air:

$3.30(O_{2} moles)*\frac{0.79(N_{2} moles)}{0.21(O_{2} moles)}=121.414 (N_{2} moles )$

Then calculate the oxygen moles number leaving the reactor, considering that 0.85 ethanol moles react and the stoichiometry of the reaction:

$3.30(O_{2} moles)-0.85(etOHmoles)*\frac{3(O_{2} moles)}{1(etOHmoles)} =0.75O_{2} moles$

Calculate the number of moles of CO2 and water considering the same:

$0.85(etOHmoles)*\frac{3(H_{2}Omoles)}{1(etOHmoles)}=2.55(H_{2}Omoles)$

$0.85(etOHmoles)*\frac{2(CO_{2}moles)}{1(etOHmoles)}=1.7(CO_{2}moles)$

The total number of moles at the reactor output would be:

$N=1.7(CO2)+12.414(N2)+2.55(H2O)+0.75(O2)\\ N=17.414(Dry-air-moles)$

So, the oxygen mole fraction would be:

$y_{O_{2}}=\frac{0.75}{17.414}=0.0430=4.3$ %

6 0

3 years ago

What is the molar mass of water in the hydrate? FeSO4 • 7H2O

Rainbow [258]

M=7M(H₂O)

M=7*18.015 g/mol = 126.105 g/mol

5 0

3 years ago

What is the expected oxidation state for the most common ion of element 2

lozanna [386]

Answer: 1+

Justification:

The ionization energies tell the amount of energy needed to release an electron and form a ion. The first ionization energy if to loose one electron and form the ion with oxidation state 1+, the second ionization energy is the energy to loose a second electron and form the ion with oxidation state 2+, the third ionization energy is the energy to loose a third electron and form the ion with oxidation state 3+.

The low first ionization energy of element 2 shows it will lose an electron relatively easily to form the ion with oxidations state 1+.

The relatively high second ionization energy (and third too) shows that it is very difficult for this atom to loose a second electron, so it will not form an ions with oxidation state 2+. Furthermore, given the relatively high second and third ionization energies, you should think that the oxidation states 2+ and 3+ for element 2 never occurs.

Therefore, the expected oxidation state for the most common ion of element 2 is 1+.

3 0

2 years ago