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3 years ago

Help me cuz it's due in 5 minutes!!!!

Chemistry

1 answer:

Katyanochek1 [597]3 years ago

6 0

115 is the correct answers for your questions

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Complete the equation C14H30 = ________ + C7H14

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Answer:

C14H30 -----> C7H14 + C2H4 +C5H10

Explanation:

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During a 4 hour trip, a boat sailed 70 km in 2 h. Then the boat did not move for 1 hour and traveled 26 km in the last hour. Wha

Lubov Fominskaja [6]

Answer:

32km/hr

Explanation:

The question asks to calculate the average speed for a boat.

Mathematically, the average speed is the total distance divided by the total time

Let’s get the total distance traveled:

That would be 70km + 26km = 96km

Now the total time taken would be 3 as the boat did not move at all for an hour.

The average speed is thus 96/3 = 32km/hr

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3 years ago

Calculate the standard reaction Gibbs free energy for the following cell reactions: (a) 2 Ce41(aq) 1 3 I2(aq) S 2 Ce31(aq) 1 I32

Law Incorporation [45]

Answer:

For a: The standard Gibbs free energy of the reaction is -347.4 kJ

For b: The standard Gibbs free energy of the reaction is 746.91 kJ

Explanation:

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$ ............(1)

For a:

The given chemical equation follows:

$2Ce^{4+}(aq.)+3I^{-}(aq.)\rightarrow 2Ce^{3+}(aq.)+I_3^-(aq.)$

Oxidation half reaction: $Ce^{4+}(aq.)\rightarrow Ce^{3+}(aq.)+e^-$ ( × 2)

Reduction half reaction: $3I^_(aq.)+2e^-\rightarrow I_3^-(aq.)$

We are given:

$n=2\\E^o_{cell}=+1.08V\\F=96500$

Putting values in equation 1, we get:

$\Delta G^o=-2\times 96500\times (+1.80)=-347,400J=-347.4kJ$

Hence, the standard Gibbs free energy of the reaction is -347.4 kJ

For b:

The given chemical equation follows:

$6Fe^{3+}(aq.)+2Cr^{3+}+7H_2O(l)(aq.)\rightarrow 6Fe^{2+}(aq.)+Cr_2O_7^{2-}(aq.)+14H^+(aq.)$

Oxidation half reaction: $Fe^{3+}(aq.)\rightarrow Fe^{2+}(aq.)+e^-$ ( × 6)

Reduction half reaction: $2Cr^{2+}(aq.)+7H_2O(l)+6e^-\rightarrow Cr_2O_7^{2-}(aq.)+14H^+(aq.)$

We are given:

$n=6\\E^o_{cell}=-1.29V\\F=96500$

Putting values in equation 1, we get:

$\Delta G^o=-6\times 96500\times (-1.29)=746,910J=746.91kJ$

Hence, the standard Gibbs free energy of the reaction is 746.91 kJ

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3 years ago

2. An electric heating element is connected to a 110 V circuit and a current of 3.2 A is flowing though the element. How much en

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Help me cuz it's due in 5 minutes!!!! ​

Help me cuz it's due in 5 minutes!!!!