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Soloha48 [4]
3 years ago
10

Need help with matching the pictures with biomes

Chemistry
2 answers:
vitfil [10]3 years ago
7 0

Answer:

what picture

Explanation:

b DH dhdhdhdh

never [62]3 years ago
6 0
There’s no imageee attached
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What are the common units for expressing solution concentration?
Ira Lisetskai [31]
It would be mol/dm^3 Formula of concentration: (no. of moles/volume of solution in dm^3) 1 cm^3 = 0.001dm^3 Hope this helped. :)
6 0
3 years ago
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Samira has a son who is 24 months old. Her sister Paulina also has a son who is one and a half years old. Whose son is older? Ex
Troyanec [42]
Samira’s son is older because there are 12 months in a year and we can gather that 12+12=24 so we know that Paulina’s son is younger because he is only 18 months.
8 0
3 years ago
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Which of the following is the major advantage of using geothermal energy instead of using coal energy? Choice:
maxonik [38]

Answer:

C. Releases less carbon dioxide

Explanation:

since geothermal steams are directly from earths surface they do not have carbon dioxide therefore they can be used to rotate turbines ultimately generating electricity.

5 0
3 years ago
Which of the procedures, if either, is more accurate when making a 1/501/50 dilution of a solution? Transfer 1 mL1 mL with a pip
nikitadnepr [17]

Full question:

First of all, bear in mind that this question is incomplete. Here is the full question:

  • Which of the procedures, if either, is more accurate when making a 1/50 dilution of a solution?

a) Transfer 1 mL with a pipet into a 50-mL volumetric flask.

b) Transfer 20 mL with a pipet into a 1-L volumetric flask.

c) Both procedures have the same accuracy.

  • How can the accuracy of either procedure be improved?

a) Use an Erlenmeyer flask instead of a volumetric flask.

b) Use a graduated cylinder instead of a pipet for the transfer.

c) Calibrate each piece of glassware.

d) Instead of using a 1/50 dilution to make the solution, weigh out the material on a balance and transfer ir directly to the volumetric flask.

Answer:

1) c) Both procedures have the same accuracy.

2) c) Calibrate each piece of glassware.

Explanation:

1) Given that the 1/50 relation is maintained whether we pour 1 mL into a 50-mL volumetric flask or we transfer 20 mL with a pipet into a 1-L volumetric flask, both procedures have the same accuracy (because in both procedures we are using volumetric glassware).

When we want the most exact result possible, we need to calibrate the volumetric glassware used (bear in mind that the only volumetric glassware is the buret, the volumetric flask, the micropipet and the pipet). This is usually done measuring the mass of water poured by the recipient or contained in it, and using the density of that liquid to convert mass into volume. In this way is possible, for example, to determine that a pipet poured 10.016 mL and not 10.000 mL.

3 0
3 years ago
How to solve for K when given your anode and cathode equations and voltage
Aleks04 [339]

Answer:

See Explanation

Explanation:

In thermodynamics theory the Free Energy (ΔG) of a chemical system is described by the expression ΔG = ΔG° + RTlnQ. When chemical system is at equilibrium ΔG = 0. Substituting into the system expression gives ...

0 = ΔG° + RTlnKc, which rearranges to ΔG° = - RTlnKc.  ΔG° in electrochemical terms gives ΔG° = - nFE°, where n = charge transfer, F = Faraday Constant = 96,500 amp·sec and E° = Standard Reduction Potential of the electrochemical system of interest.

Substituting into the ΔG° expression above gives

-nFE°(cell) = -RTlnKc => E°(cell) = (-RT/-nF)lnKc = (2.303·R·T/n·F)logKc

=> E°(cell) = (0.0592/n)logKc = E°(Reduction) - E°(Oxidation)

Application example:

Calculate the Kc value for a Zinc/Copper electrochemical cell.

Zn° => Zn⁺² + 2e⁻  ;    E°(Zn) = -0.76 volt  

Cu° => Cu⁺² + 2e⁻ ;    E°(Cu) =  0.34 volt

By natural process, charge transfer occurs from the more negative reduction potential to the more positive reduction potential.

That is,

           Zn° => Zn⁺² + 2e⁻ (Oxidation Rxn)

Cu⁺² + 2e⁻ => Cu°             (Reduction Rxn)

E°(Zn/Cu) = (0.0592/n)logKc

= (0.0592/2)logKc = E°(Cu) - E°(Zn) = 0.34v - (-0.76v) = 1.10v

=> logKc = 2(1.10)/0.0592 = 37.2

=> Kc = 10³⁷°² = 1.45 x 10³⁷

3 0
3 years ago
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