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mr_godi [17]
2 years ago
12

What is the relationship between the electron configuration of an element and the period in which that element appears in the pe

riodic table?.
Chemistry
1 answer:
MrMuchimi2 years ago
4 0

Answer:

The links are: the number of circles in the electronic configuration of an element is represented in the periodic table as the period number that element is situated in. the number of electrons in the outermost shell of an element is represented in the periodic table as the group number that element is situated in.

Explanation:

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The total number of oxygen atoms indicated by formula Cu3(PO4)2?
ASHA 777 [7]

Answer:

There's a total of 8 oxygen atoms in the formula Cu3(PO4)2

6 0
3 years ago
I NEED THE ANSWER QUICKKK!!!
julsineya [31]

Answer:  This is hard to do accurately, but here is my best assessment.

Explanation:

Experiment:  B  -  describes how the experiment was done

Conclusion:  A  -  The data support the prevailing hypothesis

Research:  D   -   This is what we analyzed

Analysis:  C  -  We compared the data

7 0
2 years ago
Milk of magnesia has the chemical formula Mg(OH)2. What is the mass of 3.2 moles of milk of magnesia? The molar Mass of magnesiu
GenaCL600 [577]
To calculate the mass of milk of magnesia given, we need certain data like molar mass of the compound which needs the atomic mass of the atoms in the compound. We calculate as follows:

Molar mass of <span>Mg(OH)2 = 24.3 g/mol + (2 x (16 + 1.0)) = 58.30 g/mol

Mass = 3.2 mol (</span>58.30 g/mol) = 186.56 grams
3 0
3 years ago
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
19. If it takes more energy to break the bonds of the reactants than is released when the bonds of
frozen [14]

Answer:

If it takes more energy to break the original bonds than is released when the new bonds are formed, then the net energy of the reaction is negative. This means that energy must be pumped into the system to keep the reaction going. Such reactions are known as endothermic.

Explanation:

5 0
3 years ago
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