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In the following reaction, how many grams of oxygen will react with 10.47 grams of benzene (C6H6)? 2C6H6 + 1502 12CO2 + 6H2O The

goldfiish [28.3K]

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8 0

3 years ago

In an experiment, a 0.5297 g sample of diphenylacetylene (C14H10) is burned completely in a bomb calorimeter. The calorimeter is

V125BC [204]

Answer:

the Molar heat of Combustion of diphenylacetylene $(C_{14}H_{10})$ = $-6.931 *10^3 \ kJ/mol$

Explanation:

Given that:

mass of diphenylacetylene $(C_{14}H_{10})$ = 0.5297 g

Molar Mass of diphenylacetylene $(C_{14}H_{10})$ = 178.21 g/mol

Then number of moles of diphenylacetylene $(C_{14}H_{10})$ = $\frac{mass}{molar \ mass}$

= $\frac{0.5297 \ g }{178.24 \ g/mol}$

= 0.002972 mol

By applying the law of calorimeter;

Heat liberated by 0.002972 mole of diphenylacetylene $(C_{14}H_{10})$ = Heat absorbed by $H_2O$ + Heat absorbed by the calorimeter

Heat liberated by 0.002972 mole of diphenylacetylene $(C_{14}H_{10})$ = msΔT + cΔT

= 1369 g × 4.184 J g⁻¹°C⁻¹ × (26.05 - 22.95)°C + 916.9 J/°C (26.05 - 22.95)°C

= 17756.48 J + 2842.39 J

= 20598.87 J

Heat liberated by 0.002972 mole of diphenylacetylene $(C_{14}H_{10})$ = 20598.87 J

Heat liberated by 1 mole of diphenylacetylene $(C_{14}H_{10})$ will be = $\frac{20598.87 \ J}{0.002972 \ mol}$

= 6930979.139 J/mol

= 6930.98 kJ/mol

Since heat is liberated ; Then, the Molar heat of Combustion of diphenylacetylene $(C_{14}H_{10})$ = $-6.931 *10^3 \ kJ/mol$

3 0

3 years ago

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Below are the reduction half reactions for chemolithoautotrophic nitrification, where ammonia is a source of electrons and energ

SSSSS [86.1K]

Answer: The molecules of oxygen gas that will be reduced to water are 42 molecules

Explanation:

We are given:

$E^o_{(NO_2^-/NH_4)}=-0.41V\\E^o_{(O_2/H_2O)}=0.82V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, oxygen will undergo reduction reaction will get reduced.

$NH_4$ will undergo oxidation reaction and will get oxidized.

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

The half reactions follows:

Oxidation half reaction: $NH_4\rightarrow NO_2^-+6e^-$ ( × 4)

Reduction half reaction: $O_2+4e^-\rightarrow 2H_2O$ ( × 6)

Overall reaction: $4NH_4+6O_2\rightarrow 4NO_2^-+12H_2O$

We are given:

Molecules of $NH_4$ = 28

By Stoichiometry of the reaction:

4 molecules of $NH_4$ reacts with 6 molecules of oxygen gas

So, 28 molecules of $NH_4$ will react with = $\frac{6}{4}\times 28=42$ molecules of oxygen gas

Hence, the molecules of oxygen gas that will be reduced to water are 42 molecules

4 0

3 years ago

PLEASE HELP

Sergio [31]

Answer:

C. The Protons

5 0

3 years ago

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Which gas law (choices are Charles' Law, Gay-Lussac's Law, or Boyle's Law) explains each scenario:________.

posledela

Answer:

A. Boyle's Law

B. Charles' Law

C. Gay-Lussac's Law

Explanation:

An air bag inflates due to the decomposition of sodium azide or NaN₃ to completely fill the bag with nitrogen gas which is an example of Boyle's law, which states that the pressure of a given mass of gas is inversely proportional to its volume, hence due to the estricted volume of the airbag, the pressure of the nitrogen gas in the bag increses protecting the occupants of a cr from injuries in a crash

Helium balloon decrease in sice in a freezer is an example of Charlles law which states that the volume of a given mass of gas is nverslely proportionl to its temperature at constant pressure

A can of spray paint will explode if tossed into a fire is an example of Gay-Lussac's Law which states that the pressure of a given mass of gas is directly proportional to its temperature hence the increased pressure causes the can ti explode

4 0

2 years ago