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Vitek1552 [10]
3 years ago
15

Situation 1

Chemistry
1 answer:
SIZIF [17.4K]3 years ago
8 0

Answer:

3. 17

4. 18

5. 10

6. Cation

7. 35

8. 81

9. 36

10. Anion

11. 37

12. 48

13. +1

14. Cation

15. 5

16. 5

17. 2

18. Cation

Explanation:

Situation 1

The mass number = 35 = The sum of the protons and the neutrons in an atom

The number of protons = 17

The net charge = 1

3. The atomic  number = The number of proton = 17

4. The number of neutrons = Mass number - Atomic number

∴ The number of neutrons = 35 - 17 = 18

5. Given that the net charge of the atom = +1, the number of electrons in the atom is 1 less than the number of protons

Therefore, the number of electrons = 11 - 1 = 10

6. Positively charged ions are called a cation

The atom is a positively charged and it is therefore it is a cation

Situation 2

The number of protons =  35

The number of neutrons = 46

The net charge = -1

7. The atomic number = The number of protons = 35

8. The mass number = The number of protons + The number of neutrons

∴ The mass number = 35 + 46 = 81

9. The number of electrons = The number of protons + 1

∴ The number of electrons = 35 + 1 = 36

10. The type of atom is an anion

Situation 3

Atomic number = 37

Mass number = 85

Number of electrons = 36

11. The number of protons = The atomic number = 37

12. The number of neutrons = The mass number - The atomic number

∴ The number of neutrons = 85 - 37 = 48

13. The net charge = The atomic number - The number of electrons

∴ The net charge = 37 - 36 = +1

14. Given that the net charge = +1, the type of atom is a cation

Situation 4

Mass number = 11

Number of neutrons = 6

Net charge = +3

15. The atomic number = The mass number - The number of neutrons

∴ The atomic number = 11 - 6 = 5

16. The number of protons = The atomic number = 5

17. Given that the net charge is +3, we have;

The number of electrons = The number of protons - The net charge

∴ The number of electrons = 5 - (+3) = 2

18. The type of atom with a net positive charge (+3) is a cation

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0. 414

Explanation:

Octahedral interstitial lattice sites.

Octahedral interstitial lattice sites are in a plane parallel to the base plane between two compact planes and project to the center of an elementary triangle of the base plane.

The octahedral sites are located halfway between the two planes. They are vertical to the locations of the spheres of a possible plane. There are, therefore, as many octahedral sites as there are atoms in a compact network.

The Octahedral interstitial void ratio range is 0.414 to 0.732. Thus, the minimum cation-to-anion radius ratio for an octahedral interstitial lattice site is 0. 414.

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It breaks down into multiple elements

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The pH of the solution : 12

<h3>Further explanation</h3>

Reaction

HCOOH    +     NaOH   ⇒     HCOONa   +   H₂O

mol HCOOH =

\tt 20~ml\times 0.2~M=4~mlmol

mol NaOH =

\tt 25~ml\times 0.2~M=5~mlmol

Mol NaOH>mol HCOOH ⇒ at the end of the reaction there will be a strong base remains from mol NaOH, so that the pH is determined from [OH⁻]

ICE method :

HCOOH    +     NaOH   ⇒     HCOONa   +   H₂O

4                          5

4                          4                     4                   4

0                          1                      1                    1

Concentration of [OH⁻] from NaOH :

\tt \dfrac{1~mlmol}{20+25~ml}=0.02

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3 years ago
What is the empirical formula of an oxide of nitrogen containing 63.61% by mass of nitrogen and 36.69% by mass of oxygen?
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N₂O is the empirical formula of an oxide of nitrogen containing 63.61% by mass of nitrogen and 36.69% by mass of oxygen.

Empirical formula can be calculated by

Suppose we have 100 g of the substance. That indicates that it has 36.69 grams of oxygen and 63.61 grams of nitrogen.

Masses transformed into moles:

Formula used

Given mass/ Molar mass

14.01 g contains 1 mol of N

So 63.61 g of N contains moles is equals to

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Similarly

16 g of O contains 1 mole of O

36.69 g of O contains moles is equals to

(1 mol O / 16.00 g O) 36.69 g O = 2.293 mol O

Divide by the smallest to normalize:

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Therefore, there are roughly twice as many N as O atoms. N2O is the empirical formula as a result.

Ratio is basically 2:1

Hence, N₂O is the empirical formula of an oxide of nitrogen

Learn more about Empirical Formula here brainly.com/question/27873410

#SPJ4

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