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2 years ago

Help me please Can someone help me and explain how to find the configuration pleasee

Chemistry

1 answer:

iogann1982 [59]2 years ago

3 0

The numbers in the powers are the number of electrons in a specific orbital. so you can add them to find the atomic number

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In a thin layer chromatography experiment, a plate of length 9.3 cm was used and a horizontal line was made at 1.45 cm above the

dezoksy [38]

Answer: The $R_f$ value is 0.664

Explanation:

Distance travelled by solvent front = (7.7-1.45)cm = 6.25 cm

Distance travelled by unknown = (5.6-1.45) cm = 4.15 cm

The retention factor or the $R_f$ value is defined as the ratio of distance traveled by the unknown to the distance traveled by the solvent front.

$R_f=\frac{\text {distance travelled by unknown}}{\text {distance travelled by solvent}}$

$R_f=\frac{4.15}{6.25}=0.664$

Thus the $R_f$ value is 0.664

6 0

2 years ago

Where is a product found at the beginning or end of a chemical reaction?

olya-2409 [2.1K]

A reaction is when two or more pure substances combine with each other to form another identity of pure substances. In general from, it is written as:

A + B → C + D

The substances A and B are the reactants, while the substances C and D are the products. Therefore, in a reaction, the products are found at the end or right side of the reaction.

3 0

3 years ago

How many moles of electrons are required to reduce one mole of bromine gas (Br2) to two moles of bromide ions (Br-)? 1 e- 2 e- 3

rosijanka [135]

The correct answer would be the second option. It would would need 2 moles of electrons to reduce one mole of bromine gas into two moles of the bromide ions. This is a reduction reaction. It would be written as:

Br2 = 2Br- + 2e-

8 0

2 years ago

quien y en que cantidad sera el reactivo limite, si utilizamos 125 g de ácido, H (CH3COO) y 275 g de hidróxido , Al(OH)3

Deffense [45]

Answer:

The limiting reactant is acetic acid. All 125 g will react.

Explanation:

1. Assemble the information

We will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.

Mᵣ: 60.05 78.00

3CH₃COO-H + Al(OH)₃ ⟶ (CH₃COO)₃Al + 3H₂O

Mass/g: 125 275

2. Calculate the moles of each reactant

$\text{Moles of CH$_{3}$COOH} = \text{125 g CH$_{3}$COOH} \times \dfrac{\text{1 mol CH$_{3}$COOH}}{\text{60.05 g CH$_{3}$COOH}} = \text{2.082 mol CH$_{3}$COOH }\\\\\text{Moles of Al(OH)}_{3} = \text{275 g Al(OH)}_{3} \times \dfrac{\text{1 mol Al(OH)}_{3}}{\text{78.00 g Al(OH)}_{3}} = \text{3.526 mol Al(OH)}_{3}$

3. Calculate the moles of (CH₃COO)₃Al from each reactant

$\textbf{From CH$_{3}$COOH:}\\\text{Moles of (CH$_{3}$COO)$_{3}$Al} = \text{2.082 mol CH$_{3}$COOH} \times \dfrac{\text{1 mol (CH$_{3}$COO)$_{3}$Al}}{\text{3 mol CH$_{3}$COOH}}\\\\= \text{0.6939 mol (CH$_{3}$COO)$_{3}$)Al}\\\textbf{From Al(OH)}_{3}:\\\text{Moles of (CH$_{3}$COO)$_{3}$Al } =\text{3.526 Al(OH)}_{3} \times \dfrac{\text{1 mol (CH$_{3}$COO)$_{3}$Al }}{\text{1 mol Al(OH)}_{3}}\\\\= \text{3.526 mol (CH$_{3}$COO)$_{3}$Al}$

$\text{Acetic acid is the $\textbf{limiting reactant}$ because it gives fewer moles of} \\\text{(CH$_{3}$COO)$_{3}$Al. All $\textbf{125 g}$ will react.}$

5 0

2 years ago

Most acids form ions in solution by ionizing, and most bases form ions in solution by

Fiesta28 [93]

There are other ions that make acidic and basic solutions, but we won't be talking about them here. That pH scale we talked about is actually a measure of the number of H + ions in a solution. If there are a lot of H + ions, the pH is very low.

Hope I helped :)

8 0

2 years ago