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Marianna [84]
3 years ago
5

A) ambos pertencem ao 5o período da Tabela Periódica;

Chemistry
1 answer:
Reptile [31]3 years ago
8 0

Answer:

A

Explanation:

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Carbon exists in nature as three isotopes. Given the atomic masses and relative abundance of the isotopes, what is the average a
Ivan

The most abundant carbon isotope is carbon-12.

The relative atomic mass of carbon is 12.011, which is extremely close to 12.0. This means that the masses C-13, and C-14 are practically negligible when contributing to the relative atomic mass of carbon.

the C-12 isotope makes up 98.9% of carbon atoms, C-13 makes up 1.1% of carbon atoms, and C-14 makes up just a trace of carbon atoms as they are found in nature.

6 0
3 years ago
Magnesium chloride is a salt formed with ionic bonds between one magnesium ion and two chloride ions. Magnesium has two electron
katrin [286]

Answer:

The magnesium atom loses 2 electron to the 2 atoms of chlorine. The 7 valency electrons of each chlorine atom will now be 8 to attain stable configuration. The final compound is written as MgCl2.

Explanation:

Ionic compounds are compound formed from the transfer of electron(s). One atom of the element loses electron(s) while the other atom gains electron(s).

The compound Magnesium chloride is an ionic compound . The bond between an atom of magnesium and 2 atoms of chlorine is an ionic bonding.

The valency electron of magnesium is 2 electron , for the atom of magnesium to  attain octet rule, it will easily lose it 2 electrons to the chlorine atoms.

The chlorine atom on the other hand has 7 valency electrons, to attain octet configuration it will most likely gain 1 electron to become stable.

The magnesium atom loses 2 electron to the 2 atoms of chlorine. The 7 valency electrons of each chlorine atom will now be 8 to attain stable configuration. The final compound is written as MgCl2.

7 0
3 years ago
1. Interpret the following equation using moles, molecules, and volumes (assume STP). Compare the mass of the reactants to the m
blsea [12.9K]
1. In this reaction, 2 moles of nitrogen gas reacts with 3 moles of oxygen gas to give 2 moles of N2O3 gas. 2 nitrogen molecules react with 3 oxygen molecules to give 2 N2O3 molecules.  Under STP, one mole of an ideal gas occupies a volume of 22.4 liters. So in this reaction, 44.8 liters of nitrogen gas reacts with 67.2 liters of oxygen gas to give 44.8 liters of N2O3 gas.  The total mass of the reactants (N2 and O2) is the same as the total mass of the product (N2O3). This is called mass balance of a chemical reaction.


2. According to the chemical reaction, 3 moles of chlorine gas produces 2 moles of iron(III) chloride.  So, to produce 1 moles of iron(III) chloride, 3/2 (1.5) moles of chlorine gas is required.  Therefore, to produce 14 moles of iron(III) chloride, 14 x 1.5 = 21 moles of chlorine is needed.
3 0
3 years ago
Read 2 more answers
Calculate the empirical formula for each stimulant based on its elemental mass percent composition. a. nicotine (found in tobacc
Ira Lisetskai [31]

This an incomplete question, here is a complete question.

Calculate the empirical formula for each stimulant based on its elemental mass percent composition.

a. nicotine (found in tobacco leaves): C 74.03%, H 8.70%, N 17.27%

b. caffeine (found in coffee beans): C 49.48%, H 5.19 %, N 28.85% and O 16.48%

Answer:

(a) The empirical formula for the given compound is C_5H_7N

(b) The empirical formula for the given compound is C_4H_5N_2O

Explanation:

<u>Part A: nicotine </u>

We are given:

Percentage of C = 74.03 %

Percentage of H = 8.70 %

Percentage of N = 17.27 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 74.03 g

Mass of H = 8.70 g

Mass of N = 17.27 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{74.03g}{12g/mole}=6.17moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{8.70g}{1g/mole}=8.70moles

Moles of Nitrogen = \frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{17.27g}{14g/mole}=1.23moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.23 moles.

For Carbon = \frac{6.17}{1.23}=5.01\approx 5

For Hydrogen  = \frac{8.70}{1.23}=7.07\approx 7

For Nitrogen = \frac{1.23}{1.23}=1

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N = 5 : 7 : 1

The empirical formula for the given compound is C_5H_7N_1=C_5H_7N

<u>Part B: caffeine</u>

We are given:

Percentage of C = 49.48 %

Percentage of H = 5.19 %

Percentage of N = 28.85 %

Percentage of O = 16.48 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 49.48 g

Mass of H = 5.19 g

Mass of N = 28.85 g

Mass of O = 16.48 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = \frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{49.48g}{12g/mole}=4.12moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{5.19g}{1g/mole}=5.19moles

Moles of Nitrogen = \frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{28.85g}{14g/mole}=2.06moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{16.48g}{16g/mole}=1.03moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.03 moles.

For Carbon = \frac{4.12}{1.03}=4

For Hydrogen  = \frac{5.19}{1.03}=5.03\approx 5

For Nitrogen = \frac{2.06}{1.03}=2

For Nitrogen = \frac{1.03}{1.03}=1

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N : O = 4 : 5 : 2 : 1

The empirical formula for the given compound is C_4H_5N_2O_1=C_4H_5N_2O

6 0
3 years ago
A particular uranium alloy has a density of 18.75 g/cm3.
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