To cook a steak to medium-rare, it needs to have an internal temperature of 135

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:

Part A: nicotine

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 $C_5H_7N_1=C_5H_7N$

Part B: caffeine

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

I went for a walk the other day. I went four blocks east, then seven blocks south, then one block west and finally

nalin [4]

Answer:

a) distance is 4+7+1+8=20 blocks

b) displacement is 10 blocks

Explanation:

find displacement: x and y

x axis displacement = 4-1 = 3 blocks

y axis displacement = -7+8= 1 block

displacement = the square root of 3^2 + 1^2

= 9+1 = 10 blocks.

You can find the angle of displacement with respect to the initial position using trig identities, if you wish.

4 0

2 years ago

Chemistry, please help! Thanks.

Ilya [14]

Answer:

d. The gold(III) ion is most easily reduced.

Explanation:

The standard reduction potentials are

Au³⁺ + 3e⁻ ⟶ Au; 1.50 V

Hg²⁺ + 2e⁻ ⟶ Hg; 0.85 V

Zn²⁺ + 2e⁻ ⟶ Zn; -0.76 V

Na⁺ + e⁻ ⟶ Na; -2.71 V

A more positive voltage means that there is a stronger driving force for the reaction.

Thus, Au³⁺ is the best acceptor of electrons.

Reduction Is Gain of electrons and, Au³⁺ is gaining electrons, so

Au³⁺ is most easily reduced.

4 0

3 years ago

Which of the following is not a step in balancing redox reactions in acidic solution, using the half-reaction method?

schepotkina [342]

B Divide the chemical equation into two half-reaction equations, identifying which half-reaction is oxidation and which is reduction

7 0

3 years ago

Read 2 more answers

Last week you reacted magnesium with a hydrochloric acid aqueous solution and hydrogen gas was produced. Let's say that you coll

miv72 [106K]

Answer:

69.8 kilo Pasacl is the pressure of the hydrogen gas.

Explanation:

$Mg+2HCl\rightarrow MgCl_2+H_2$

Pressure at which hydrogen gas collected = p = 101.2 kilo Pascals

Vapor pressure water = $p^o$ = 31.4 kilo Pascals

The pressure of hydrogen gas = P

The pressure at which gas was collected was sum of vapor pressure of water and hydrogen gas.

$p=P+p^o$

$P =p-p^o=101.2 kPa-31.4 kPa=69.8 kPa$

69.8 kilo Pasacl is the pressure of the hydrogen gas.

3 0

3 years ago