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ArbitrLikvidat [17]
1 year ago
12

Never mind, question solved.

Chemistry
1 answer:
Alchen [17]1 year ago
5 0
The answer to the problem is 100
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This would be measurements.
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Explain why the following chemical equation represents a Lewis acid-base reaction. H+ + NH3 —>NH4+
lozanna [386]

Answer:

Due to an electron-pair acceptor and donor.

Explanations:

<em><u>Lewis acid</u></em> can be defined as an electron-pair acceptor. An example is Hydrogen ion(H+). This is because it is a proton and it distributes positive charge which means that it accepts electrons(negative charge).

<em><u>Lewis base</u></em> can be defined as an electron-pair donor. This is because it donates electrons to be accepted by the proton. An example is ammonia(NH3).

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3 years ago
Consider a generic redox reaction?
vesna_86 [32]
The Nernst equation allows us to predict the cell potential for voltaic cells under conditions other than the standard conditions of 1M, 1 atm, 25°C. The effects of different temperatures and concentrations may be tracked in terms of the Gibbs energy change ΔG. This free energy change depends upon the temperature & concentrations according to   ΔG = ΔG°  +  RTInQ  where ΔG° is the free energy change under conditions and Q is the thermodynamic reaction quotient. The free energy change is related to the cell potential  Ecell by ΔG= nFEcell

so for non-standard conditions
              -nFEcell = -nFE°cell  + RT InQ

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5 0
3 years ago
How many moles of methanol must be added to 4.50 kg of water to lower its freezing point to -11.0 ∘c? for each mole of solute, t
DochEvi [55]

<u>Given:</u>

Mass of solvent water = 4.50 kg

Freezing point of the solution = -11 C

Freezing point depression constant = 1.86 C/m

<u>To determine:</u>

Moles of methanol to be added

<u>Explanation:</u>

The freezing point depression ΔTf is related to the molality m through the constant kf, as follows:

ΔTf = kf*m

where ΔTf = Freezing point of pure solvent (water) - Freezing pt of solution

ΔTf = 0 C - (-11.0 C) = 11.0 C

m = molality = moles of methanol/kg of water = moles of methanol/4.50 kg

11.0 = 1.86 * moles of methanol/4.50

moles of methanol = 26.613 moles

Ans: Thus around 26.6 moles of methanol should be added to 4.50 kg of water.



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Wouldnt you find that on the grid or the table
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