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3 years ago

How many electrons are in an uncharged atom of chromium

2 answers:

5 0

Look up chromium on periodic table.
It's atomic number is 24.
It gives number of protons in an element.
Since the atom is uncharged (no extra or subtracted electrons), the number of protons will equal the number of electrons.

Aloiza [94]3 years ago

3 0

There are 24 electrons in the chromium element.

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In a particular experiment, 2.50-g samples of each reagent are reacted. The theoretical yield of lithium nitride is ________ g.

Neporo4naja [7]

Answer:

4.18 g

Explanation:

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Given: For Li

Given mass = 2.50 g

Molar mass of Li = 6.94 g/mol

Moles of Li = 2.50 g / 6.94 g/mol = 0.3602 moles

Given: For $N_2$

Given mass = 2.50 g

Molar mass of $N_2$ = 28.02 g/mol

Moles of $N_2$ = 2.50 g / 28.02 g/mol = 0.08924 moles

According to the given reaction:

$6Li+N_2\rightarrow 2Li_3N$

6 moles of Li react with 1 mole of $N_2$

1 mole of Li react with 1/6 mole of $N_2$

0.3602 mole of Li react with $\frac {1}{6}\times 0.3602$ mole of $N_2$

Moles of $N_2$ that will react = 0.06 moles

Available moles of $N_2$ = 0.08924 moles

$N_2$ is in large excess. (0.08924 > 0.06)

Limiting reagent is the one which is present in small amount. Thus,

Li is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

6 moles of Li gives 2 mole of $Li_3N$

1 mole of Li gives 2/6 mole of $Li_3N$

0.3602 mole of Li react with $\frac {2}{6}\times 0.3602$ mole of $Li_3N$

Moles of $Li_3N$ = 0.12

Molar mass of $Li_3N$ = 34.83 g/mol

Mass of $Li_3N$ = Moles × Molar mass = 0.12 × 34.83 g = 4.18 g

Theoretical yield = 4.18 g

5 0

3 years ago

Using the molecular orbital model to describe the bond- ing in F2????, F2, and F2????, predict the bond orders and the relative

masya89 [10]

Answer: F2 : bond order= 1.0

F2+: bond order = 1.5

F2- : bond order = 0.5

Explanation:

1. Starting with F2+

The configuration gives;

F2+ = 9F = 1S2.2S2.2P5

= 9F+ = 1S2.2S2.2P4 (this shows it gives out an electron)

Electronic configuration = σ 1S2σ*1S2σ2S2σ*2S2σ2Pz2 π2Px2 = π2Py2 π*2Px2 = π*2Py1

The number of Electrons = (9*2) – 1 = 18 -1 = 17

Bonding electrons (Nb) = 10

Antibonding electrons (Na) = 7

Bond order = (10-7)/2 = 3/2 = 1.5

Number of unpaired electrons = 1

2. Starting with F2

The configuration gives;

F2 = 9F = 1S2.2S2.2P5

9F = 1S2.2S2.2P5 (this shows no loss of an electron)

Electronic configuration = σ 1S2σ*1S2σ2S2σ*2S2σ2Pz2 π2Px2 = π2Py2 π*2Px2 = π*2Py2

The number of Electrons = (9*2) = 18 electrons

Bonding electrons (Nb) = 10

Antibonding electrons (Na) = 8

Bond order = (10-8)/2 = 2/2 = 1.0

Number of unpaired electrons = 0

3. Starting with F2-

The configuration gives;

F2- = 9F = 1S2.2S2.2P5

10F--= 1S2.2S2.2P6 (this shows an addition of an electron)

Electronic configuration = σ 1S2σ*1S2σ2S2σ*2S2σ2Pz2 π2Px2 = π2Py2 π*2Px2 = π*2Py2 σ*2Pz

The number of Electrons = (9*2) + 1 = 19 electrons

Bonding electrons (Nb) = 10

Antibonding electrons (Na) = 9

Bond order = (10-9)/2 = 1/2 = 0.5

Number of unpaired electrons = 1

To get the order of bond as well as length, we know that;

Bond order directly proportional to 1/ Bond length

Therefore the Ascending Bond length = F2+ ˂ F2 ˂ F2-

3 0

2 years ago

10. A 2.36-gram sample of NaHCO3 was completely decomposed in an

Airida [17]

Answer:

0.79 g

Explanation:

Let's introduce a strategy needed to solve any similar problem like this:

Apply the mass conservation law (assuming that this reaction goes 100 % to completion): the total mass of the reactants should be equal to the total mass of the products.

Based on the mass conservation law, we need to identify the reactants first. Our only reactant is sodium bicarbonate, so the total mass of the reactants is:

$m_r=m_{NaHCO_3}=2.36 g$

We have two products formed, sodium carbonate and carbonic acid. This implies that the total mass of the products is:

$m_p=m_{Na_2CO_3}+m_{H_2CO_3}$

Apply the law of mass conservation:

$m_r=m_p$

Substitute the given variables:

$m_{NaHCO_3}=m_{Na_2CO_3}+m_{H_2CO_3}$

Rearrange for the mass of carbonic acid:

$m_{H_2CO_3}=m_{NaHCO_3}-m_{Na_2CO_3}=2.36 g - 1.57 g=0.79 g$

8 0

3 years ago

6. Three ice cubes are placed inside a glass of hot chocolate. Which of the following best

ycow [4]

It’s the first 1 M yea it’s the first one

4 0

2 years ago

1. Mg2+ and Ca2+ are in the same group on the periodic table. In terms of electronic

finlep [7]

Answer:

Se detailed explanation.

Explanation:

Hello,

In this case, since both magnesium and calcium ions are in group IIA, we can review the following similar properties:

- Since both calcium and magnesium are in group IIA they have two valence electrons, it means that the both of them have two electrons at their outer shells.

- They are highly soluble in water when forming ionic bonds with nonmetals such as those belonging to halogens and oxygen's family.

- Calcium has 18 electrons and magnesium 10 which are two less than the total protons (20 and 12 respectively) since the both of them have lost two electrons due their ionized form.

- Their electron configurations are:

$Ca^{20}=1s^2,2s^2,2p^6,3s^3,3p^6,4s^2\\\\Mg^{12}=1s^2,2s^2,2p^6,3s^2$

It means that the both of them are at the $s$ region since it is the last subshell at which their electrons are.

Best regards.

8 0

3 years ago