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2 years ago

Write a complete electron configuration for borin

Chemistry

1 answer:

olga2289 [7]2 years ago

6 0

The electron configuration for Boron (B) is given as $1s^2 2s^2 2p^1$ or $[He] 2s^2 2p^1$

Answer:

Boron (B) has an atomic number equal to 5 and is an element from Group 13 of the periodic table. It has a total of 5 electrons of which 3 are valence electrons.

The electron configuration for Boron can be written with the first two electrons in the 1s orbital. Since 1s orbital can hold only two electrons, out of the remaining three electrons; two electrons will go in the 2s orbital. The remaining one electron in the 2p orbital. Therefore the Boron electron configuration will be $1s^2 2s^2 2p^1$

Since $1s^2$ is the electron configuration for Helium, Electron configuration for Boron can also be written as $[He] 2s^2 2p^1$

<em>(Refer attached figure)</em>

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In which of the following is number of electrons the ion contains given correctly? A. H+ has 1 electron. B. Br- has 34 electrons

Leviafan [203]

The correct answer would be D. Ca^2+ has 18 electrons. This is because 2+ means that 2 electrons have been removed, and Ca starts out with 20 electrons. As 20 - 2 = 18 electrons, D is the correct answer.

Hope this helps!

8 0

2 years ago

Read 2 more answers

8. Ascorbic acid (Vitamin C) cannot be stored in the body, so it must be supplied in the diet. It is a C, H, O compound. If 7.75

spin [16.1K]

Answer:

Molecular formula = C6H8O6

The molar mass of vitamin C = 176.14 g/mol

Explanation:

Step 1: Data given

Mass of vitamin C = 7.75 grams

Mass of CO2 = 11.62 grams

Mass of H2O = 3.17 grams

Molar mass of CO2 = 44.01 g/mol

Molar mass of H2O = 18.02 g/mol

Molar mass C = 12.01 g/mol

Molar mass H = 1.01 g/mol

Molar mass O = 16.0 g/mol

Step 2: Calculate moles CO2

Moles CO2 = 11.62 grams /44.01 g/mol

Moles CO2 = 0.264 moles

Step 3: Calculate moles C

For 1 mol CO2 we have 1 mol C

For 0.264 moles CO2 we have 0.264 moles

Step 4: Calculate mass C

Mass C = 0.264 moles * 12.01 g/mol

Mass C = 3.17 grams

Step 5: Calculate moles H2O

Moles H2O = 3.17 grams / 18.02 g/mol

Moles H2O = 0.176 moles

Step 6: Calculate moles H

For 1 mol H2O we have 2 moles H

For 0.176 moles H2O we have 2*0.176 = 0.352 moles

Step 6: Calculate mass H

Mass H = 0.352 moles * 1.01 g/mol

Mass H = 0.356 grams

Step 7: Calculate mass O

Mass O = 7.75 grams - 3.17 - 0.356

Mass O = 4.224 grams

Step 8: Calculate moles O

Mass O = 4.224 grams / 16.0 g/mol

Mass O = 0.264 moles

Step 9: Calculate mol ratio

We divide by the smallest amount of moles

C: 0.264 moles / 0.264 moles = 1

H: 0.352 moles / 0.264 moles = 1.33

O: 0.264 moles / 0.264 moles = 1

This means for each C atom we have 1.33 H atoms and 1 O atom

For every 3 C atoms we have 4 H atoms and 3 O atoms

The empirical formula is C3H4O3

Step 10: Calculate the molecular formula

The molecular mass of the empirical formula is 88.07

Since the molar mass of vitamin C is between 150 and 200 g/mol

We have to multiply the empirical formula by 2

Molecular formula = 2*(C3H4O3) = C6H8O6

The molar mass of vitamin C = 176.14 g/mol

8 0

3 years ago

If you had 2.50 moles of anything, how many of those things would you have?

Wittaler [7]

Answer:

1.51 x 10²⁴ things

Explanation:

According to Avogadro's Constant.

1 mole of any substance, constains 6.02×10²³ particles of that subtance.

From the question,

If we have 2.50 moles of anything,

1 mole of anything ⇒ 6.02×10²³ things

2.50 moles of anything ⇒ y things

solving for y

y = (2.50× 6.02×10²³)/1

y = 15.05×10²³

y = 1.505×10²⁴

y ≈ 1.51×10²⁴

7 0

2 years ago

Calculate the number of C, H, and O atoms in 1.50 g of glucose, a sugar

DIA [1.3K]

Chemical formula of the glucose: C₆H₁₂O₆

We calculate the molar mass:
atomic mass (C)=12 u
atomic mass (H)=1 u
atomic mass (O)=16 u

atomic weight (C₆H₁₂O₆)=6(12 u)+12(1u)+6(16 u)=72 u+12u+96 u=180 u.
Therefore : 1 mol of glucose will be 180 g
The molar mass would be: 180 g/ mol

2) we calculate the number of moles of 1.5 g.
180 g---------------------1 mol
1.5 g---------------------- x

x=(1.5 g * 1 mol) / 180 g≈8.33*10⁻³ moles

we knows that:
1 mol = 6.022 * 10²³ particles (atoms or molecules)

3)We calculate the number of molecules:

Therefore:
1 mol-----------------------6.022*10²³ molecules of glucose
8.33*10⁻³ moles-------- x

x=(8.33*10⁻³ moles * 6.022*10²³ molecules)/1 mol≈5.0183*10²¹ molecules.

4)We calculate the number of C, H and O atoms:
A molecule of glucose have 6 atoms of C, 12 atoms of H, and 6 atoms of O,
number of atoms of C=(6 atoms/1 molecule)(5.0183*10²¹molecules)≈
3.011*10²²

number of atoms of H=(12 atoms/1 molecule)(5.0183*10²¹ molecules)≈
6.022*10²² .

number of atoms of O=(6 atoms/1 molecule)(5.0183*10²¹ molecules)≈
3.011*10²²

Answer: we have 3.011*10²² atoms of C, 6.022*10²² atoms of H, and 3.011*10²² atoms of O.

4 0

3 years ago

What is always conserved in a chemical reaction?

Vaselesa [24]

The answer is b. Mass of the atoms

This is because of the law of conservation of mass. This means the mass in a system (in this case the reaction) can't change, so the quantity can't be added or removed

3 0

2 years ago