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3 years ago

15

Help me please !!!

1 answer:

lawyer [7]3 years ago

5 0

Answer:

Its B. The atoms form a bond with a bond lenth of 75 pm

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In your opinion, why do you think the atomic models are not final?

julia-pushkina [17]

Answer:

because they change

Explanation:

It was based on theories and discoveries

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2 years ago

What is the partial pressure of C?<br> atm C

s344n2d4d5 [400]

The partial pressure of carbon is 45 mm Hg.

Explanation:

The partial pressure of carbon dioxide is referred as the amount of carbon dioxide present in venous or arterial blood. It acts as a ventilation in the lungs.
There is a formula for measuring partial pressure . As we know total pressure means summation of the pressure of all the gases included .
To find partial pressure we need- total pressure* fraction of mole of that gas. The partial pressure of CO2 is more because it carries deoxygenated blood from the whole body towards the lungs.

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3 years ago

If a stone with an original velocity of 0 is falling from a ledge and takes 8 seconds to hit the ground what is the final veloci

ziro4ka [17]

Also 0 as it hits the ground and stops

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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]

<u>Answer:</u>

<u>For a:</u> The standard Gibbs free energy of the reaction is -347.4 kJ

<u>For b:</u> The standard Gibbs free energy of the reaction is 746.91 kJ

<u>Explanation:</u>

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

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

<u>For a:</u>

The given chemical equation follows:

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

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

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

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

<u>For b:</u>

The given chemical equation follows:

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

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

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

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

7 0

3 years ago

A student is investigating potential and kinetic energy by stretching a spring across a table. When the student lets go

Anna11 [10]

Answer: When the spring is recoiling

Explanation:

4 0

2 years ago

Read 2 more answers