Which of the following does NOT involve a physical change?

How many atoms are in 3.20 moles of carbon

DIA [1.3K]

Answer:

19.264× $10^{23}$ atoms are present in 3.2 moles of carbon.

Explanation:

It is known that one mole of each element is composed of Avagadro's number of atoms. This is same for all the elements in the periodic table.

So, as 1 mole of any element = Avagadro's number of atoms = 6.02× $10^{23}$ atoms

It is as simple as understanding a dozen of anything is equal to 12 pieces of that object.

As here the moles of carbon is given as 3.20 moles, the number of atoms in this mole can be determined as below.

1 mole of carbon = 6.02 × $10^{23}$ atoms

Then, 3.20 moles of carbon = 3.20 × 6.02 × $10^{23}$ atoms

Thus, 19.264× $10^{23}$ atoms are present in 3.2 moles of carbon.

6 0

3 years ago

The half-life of phosphorus-32 is 14.30 days. how many milligrams of a 20.00 mg sample of phosphorus-32 will remain after 85.80

ZanzabumX [31]

The amount of sample that is left after a certain period of time, given the half-life, h, can be calculated through the equation.

A(t) = A(o) (1/2)^(t/d)

where t is the certain period of time. Substituting the known values,

A(t) = (20 mg)(1/2)^(85.80/14.30)

Solving,

A(t) = 0.3125 mg

Hence, the answer is 0.3125 mg.

7 0

3 years ago

Gold has a molar mass of 197 g/mol. (a) how many moles of gold are in a 3.98 g sample of pure gold? (b) how many atoms are in th

Natali5045456 [20]

A.)49.4974874 moles or 49.5 moles
B.)2.980808730172671e+25 or 3e+25

6 0

3 years ago

HELpPPPPppp!!!!!! will mark brainless

ANTONII [103]

Answer:

A

Explanation:

This is because in the graph shown line A has a quite greater impact of refraction than line B .

Hence, we can conclude that line A has the greater reaction at a faster rate.

6 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