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Orlov [11]
3 years ago
13

What is the predicted electron configuration of Helium

Chemistry
1 answer:
Crank3 years ago
6 0

1s²

Explanation:

The predicted electron configuration of Helium 1s².

Helium is an element with atomic number of 2. it is located on the last group on the periodic table.

  • Electron distribution into the energy levels or sublevels of atoms can be shown using electronic configurations.

Here, the two electrons in Helium fills the first energy level with s-sublevel first.

The maximum number of electrons in the sub-level is 2: this gives the 1s² configuration.

Learn more:

Electronic configuration brainly.com/question/6312987

#learnwithBrainly

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If 16.0 grams of aluminum oxide were actually produced, what is the percent yield of the reaction below given that you start wit
Svetllana [295]

Answer: The percent of yield is 50.03%.

Explanation:

First, we need to balance the equation:

4Al + 3O_{2} ⇒ 2Al_{2}O_{3}

We need to remember that the chemical equations are written in moles, so we have to convert the amounts in grams to moles, using the molecular weight of every compound: Al2O3 (101.96 g/mol), Al (29.98 g/mol) and O2 (31.99 g/mol).

In consequence, the amounts in moles of every compound will be:

16 gAl_{2}O_{3} * \frac{1 mol}{101.96 g} =0.157 mol Al_{2}O_{3}

10 g Al * \frac{1mol}{29.98 g}= 0.333 mol Al

19 g O_{2}*\frac{1 mol}{31.99 g} =0.594 mol O_{2}

Now, we have to find out which is the limit reagent or in other words, which of the reagents will be consumed first, taking into account the stoichiometric ratio of the balanced equation:

0.333 mol Al * \frac{2 mol Al_{2}O_{3}}{4 mol Al} =0.1665 mol Al_{2}O_{3}

0.594 mol O_{2} * \frac{2 mol Al_{2}O_{3}}{3 mol O_{2}} =0.396 mol Al_{2}O_{3}

As you can see, the maximum amount (theoretically) of Al2O3 that can be produced is 0.1665 mol.

Finally, we have to use the yield formula to calculate the percent yield of the reaction:

Percent of yield = \frac{actual yield}{theoretical yield} * 100 = \frac{0.157 mol Al_{2}O_{3}}{0.1665 mol Al_{2}O_{3}} * 100 = 94.25

Therefore, the percent of yield is 50.03%.

4 0
3 years ago
Hello
nordsb [41]

Answer:

The ratio of the number of atoms of gold and silver in the ornament is 197 : 1080.

Explanation:

Let the mass of silver ornament be x.

Mass of gold polished on an ornament = 1% of x= 0.1 x

Moles of silver =\frac{x}{108 g/mol}

Number of atoms = Moles\times N_A

Where : N_A = Avogadro number

Moles of gold =\frac{0.1x}{197 g/mol}

Silver atoms =\frac{x}{108 g/mol}\times N_A

Gold atoms =\frac{0.1x}{197 g/mol}\times N_A

The ratio of the number of atoms of gold and silver in the ornament:

=\frac{\frac{0.1x}{197 g/mol}\times N_A}{\frac{x}{108 g/mol}\times N_A}

= 197 : 1080

3 0
3 years ago
How many grams of O2(g) are needed to completely burn 46.0 g C3H8(g)?
Veronika [31]

Answer:

167,3 grams of O2 (g)

Explanation:

According to the following balanced equation:

C3H8 + 5 02 ---> 3 CO2 + 4 H20

We calculate the mass of the moles:

5 mol O2 = 16grams/mol x2x 5= 160 grams/mol

1 mol C3H8= 3 x12 grams/mol + 8 x 1 gram/mol= 44 grams/mol

If 44 grams C3H8 react with--------- 160 grams O2

46 gramsC3H8 react with    --------X = (46x160)/44 =167, 3 grams O2

8 0
4 years ago
H 2(g)+I 2(g)→2 H I(g)
dlinn [17]

Answer:

The images for the 3 sub-questions have been presented in the first, second and third attached image to this solution.

Explanation:

The reaction for the reaction between Hydrogen gas and Iodine to give Hydrogen iodide

H₂ (g) + I₂ (g) → 2 H I(g)

ΔH∘rxn = −9.5 kJ/mol r x n

a) From the information provided, the heat of reaction being negative shows that the reaction is an exothermic reaction with the products for an exothermic reaction having a lower energy content/level as the reactants for this reaction.

This shows that the energy diagram will have the products at a lower level than fe reactants.

The sketch of the energy diagram for this exothermic reaction is presented in the first attached image to this solution.

The heat of reaction is shown on the energy diagram, together with the activation energy, Ea, the heat content/energy level for the reactants (Hr) & products (Hr) and finally the ΔH∘rxn.

ΔH∘rxn = Hp - Hr

b) When a catalyst that speed up a chemical reaction is introduced to the chemical reaction, the catalyst goes about making the reaction faster by lowering the activation energy of the reaction.

The activation energy is the minimum energy that the reactants must possess to the able to form products. On the graph, it is denoted as Ea, the difference in energy between the peak of the curve and the emergy level of the reactants.

So, a catalyst will cause this activation energy to be lowered to a new level of Ea₁.

The energy diagram of this is presented in the second attached image to this answer.

The ΔH∘rxn remains unchanged for this introduction of catalyst.

The broken lines are used to represent the lowered activation energy action of the catalyst.

c) The reverse reaction will have opposite the energy properties of the initial reaction.

The initially exothermic reaction with higher heat content of reactants than products is now an endothermic reaction with heat content/energy level of the products higher than that of the reactants.

The energy diagram is presented in the third attached image.

The activation energy is bigger and the ΔH∘rxn is now positive and equal to 9.5 kJ/mol, the direct additive inverse of the ΔH∘rxn for the initial exothermic reaction.

Hope this Helps!!!

7 0
3 years ago
A naturally occurring element consists of three isotopes. The data for the isotopes are:
max2010maxim [7]
(46.972* 69.472% + 48.961*21.667% + <span>49.954*8.8610%)/100% =

</span>= 46.972* 0.69472 + 48.961*0.21667 + 49.954*0.088610 =47.667 u
7 0
3 years ago
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