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2 years ago

.....................?

Chemistry

1 answer:

liraira [26]2 years ago

4 0

Answer:

Look at the periodic table of elements

Explanation:

Group 1 is the most reactive and to the right

Group 18 is least reactive and all the way to the left. The rest are in between. The groups go vertical.

You'd be able to solve with this information youre welcome

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Someone please notice!! Explain the other parts such as the petal and receptacle do in the flower.

Troyanec [42]

Answer: well the receptacle connect the stalk to the flower and to support the flower and keeps the flower in an elevated position so as to attract the insects

Explanation: I don’t know if this helped

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2 years ago

If a block has the mass of 15.00 grams and measurements of 5.00cm by 3.00cm by 2.00cm what is the volume of the block?

Natalija [7]

Answer:

Volume = 30cm³

Explanation:

A block is a geometrical figure and its volume, -look at the figure-, follows the equation:

Volume = Width*Length*Height

As the measurements of the block are 5.00cm, 3.00cm and 2.00cm, the volume is:

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2 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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Vera_Pavlovna [14]

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