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4 years ago

Which of these are true for the reaction above?

Chemistry

1 answer:

agasfer [191]4 years ago

6 0

Answer:

a, d, f

Explanation:

ΔHrxn = ΔH(CCl4) -ΔH(CH4) = - 106.7 -(-74.8) = - 31.9 kJ/mol

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There are two naturally occurring isotopes of copper. 63cu has a mass of 62.9296 amu. 65cu has a mass of 64.9278 amu. determine

SSSSS [86.1K]

1) You need to use the atomic mass of copper.

You can find it in a periodic table. It is 63.546 amu.

2) The atomic mass is the weigthed mass of the different isotopes.

This is, the atomic mass of one element is the atomic mass of each isotope times its corresponding abundance:

=> atomic mass of the element = abundance isotope 1 * atomic mass isotope 1 + abundance isotope 2 * atomic mass isotope 2 + ....+abundance isotope n * atomic mass isotope n.

3) The statement tells there are two isotopes so the abundance of one is x and the abundance of the other is 1 - x

=> 63.546 amu = x * 62.9296 amu + (1-x)*64.9278

=> 63.546 = 62.9296x + 64.9278 - 64.9278x

=> 64.9278x - 62.9296 = 64.9278 - 63.546

=> 1.9982x = 1.3818

=> x = 1.3818 / 1.9982 = 0.6915 = 69.15%

=> 1 - x = 1 - 0.6915 = 0.3085 = 30.85%

Answer:

Cu-63 69.15%;

Cu-65 : 30.85%

3 0

3 years ago

A single serving bag of snack chips contains 65.0 Cal. Assuming that all of the energy from eating these chips goes toward keepi

AveGali [126]

Answer:

= 62.1 hours

Explanation:

Energy provide by the serving is 65 cal

= 65 cal × 4.184 Kj = 271.96 kJ

271.96 KJ = 271960 J

Energy required for 1minute of energy

= 73 x 1

= 73 J/min

So, 271960 joules will be required for 271960 heart beat

Minutes = 271960 / 73

= 3593.94 minutes

Time in hours = 3725.429 / 60

= 62.1 hours

5 0

4 years ago

Benzene would react_____________.

Vlada [557]

Answer: Benzene is less reactive than methylbenzoate and more reactive than Nitrobenzene

Explanation:

This is because the methyl group on the benzene ring is an electron donating group leading to the activation of the ring and subsequently leading to more canonical resonance structure at the intermediate stage of the reaction enhancing the faster reactivity

However for the Nitrobenzene the nitro group is an electron withdrawing group leading to a slower activation and less resonance canonical structure at the reaction intermediate leading to a slower reaction than the reaction of benzene without the nitro group

4 0

4 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

Draw an electrolytic cell in which Mn2+ is reduced to Mn and Sn is oxidized to Sn2+. (Assume standard conditions).

babunello [35]

Answer:

-1.05 V

Explanation:

A detailed diagram of the setup as required in the question is shown in the image attached to this answer. The electrolytes chosen are SnCl2 for the anode half cell and MnCl2 for the cathode half cell. Tin rod and manganese rod are used as the anode and cathode materials respectively. Electrons flow from anode to cathode as indicated. The battery connected to the set up drives this non spontaneous electrolytic process.

Oxidation half equation;

Sn(s) ------> Sn^2+(aq) + 2e

Reduction half equation:

Mn^2+(aq) + 2e ----> Mn(s)

Cell voltage= E°cathode - E°anode

E°cathode= -1.19V

E°anode= -0.14 V

Cell voltage= -1.19 V - (-0.14V)

Cell voltage= -1.05 V

8 0

3 years ago