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Ludmilka [50]
3 years ago
7

Which are characteristics of all living things?

Chemistry
1 answer:
daser333 [38]3 years ago
3 0

Order, sensitivity or response to the environment, reproduction, growth and development, regulation, homeostasis, and energy processing.

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"They drink water when it is aivailable but can go weeks without it, they rely on morning dew and the water content of their food". they can use plants as resources to live.

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Help please the questions are confusing
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5- number of electrons=11

Explanation:

in a neutral atom number of protons=number of electrons which in this case=11

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What is the ph of a solution made by mixing 25.00 ml of 0.100 m hcl with 40.00 ml of 0.100 m koh? assume that the volumes of the
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<span>Answer: The HCl and KOH will react until one or the other is gone. As you have a larger volume of an equal concentration of HCl, the KOH will go first. moles HCl = 0.04000 L * 0.100 M = 0.00400 moles moles KOH = 0.02500 L * 0.100 M = 0.00250 moles moles HCl left = 0.00400 - 0.00250 = 0.00150 moles Your total volume is now 65.00 mL, so the [HCl] = 0.00150 moles / 0.06500 L = 0.0231 M = [H+] pH = -log [H+] = -log (0.0231) = 1.64</span>
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3 years ago
Determine the free energy(ΔG) from the standard cell potential (Ecell0 ) for the reaction:2ClO2-(aq)+Cl2(g)→2ClO2(g)+ 2Cl-(aq)wh
Dima020 [189]

<u>Answer:</u> The \Delta G^o for the given reaction is -7.84\times 10^4J

<u>Explanation:</u>

For the given chemical reaction:

2ClO_2^-(aq.)+Cl_2(g)\rightarrow 2ClO_2(g)+2Cl^-(aq.)

Half reactions for the given cell follows:

<u>Oxidation half reaction:</u> ClO_2^-\rightarrow ClO_2+e^-;E^o_{ClO_2^-/ClO_2}=0.954V  ( × 2)

<u>Reduction half reaction:</u> Cl_2+2e^-\rightarrow 2Cl(g);E^o_{Cl_2/2Cl^-}=1.36V

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=1.36-(0.954)=0.406V

To calculate standard Gibbs free energy, we use the equation:

\Delta G^o=-nFE^o_{cell}

Where,

n = number of electrons transferred = 2

F = Faradays constant = 96500 C

E^o_{cell} = standard cell potential = 0.406 V

Putting values in above equation, we get:

\Delta G^o=-2\times 96500\times 0.406=-78358J=-7.84\times 10^4J

Hence, the \Delta G^o for the given reaction is -7.84\times 10^4J

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