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oee [108]
3 years ago
10

Solution that is used as the titrant (put in buret) in this experiment. Substance that gives the light pink end point in this ex

periment. 5 mL of this chemical is added to the reaction when titrating the iron salt. Primary standard for this experiment. 15 mL of this solution is used for every titration in this experiment.
A. About 0.02M Potassium Permanganate
B. Oxalate ion
C. 20% Phosphoric acid
D. Manganese II ion
E. 4M Sulfuric Acid
F. Methyl Red
G. Sodium Oxalate
Chemistry
1 answer:
Alja [10]3 years ago
6 0
E is the correct answer
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7 0
3 years ago
Which conclusion was drawn from the results of
torisob [31]

Answer;

-(2) An atom is mostly empty space.

Experiment

-Rutherford conducted the "gold foil" experiment where he shot alpha particles at a thin sheet of gold. The conclusion that can be drawn from these experiment is that an atom is mostly empty space.

-Rutherford found that a small percentage of the particles were deflected, while a majority passed through the sheet. This caused Rutherford to conclude that the mass of an atom was concentrated at its center, as the tiny, dense nucleus was causing the deflections.

3 0
3 years ago
1+1 hahahahahhhahahaahahahahahahahahahahahahahaha why u dumb
-Dominant- [34]

Answer:

1+1=2 Unless this is a trick question. Then it's most likely 11.

Explanation:

4 0
2 years ago
Read 2 more answers
The standard cell potential Ec for the reduction of silver ions with elemental copper is 0.46V at 25 degrees celsius. calculate
Cloud [144]

Answer : The \Delta G for this reaction is, -88780 J/mole.

Solution :

The balanced cell reaction will be,  

Cu(s)+2Ag^+(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)

Here, magnesium (Cu) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

The half oxidation-reduction reaction will be :

Oxidation : Cu\rightarrow Cu^{2+}+2e^-

Reduction : 2Ag^++2e^-\rightarrow 2Ag

Now we have to calculate the Gibbs free energy.

Formula used :

\Delta G^o=-nFE^o

where,

\Delta G^o = Gibbs free energy = ?

n = number of electrons to balance the reaction = 2

F = Faraday constant = 96500 C/mole

E^o = standard e.m.f of cell = 0.46 V

Now put all the given values in this formula, we get the Gibbs free energy.

\Delta G^o=-(2\times 96500\times 0.46)=-88780J/mole

Therefore, the \Delta G for this reaction is, -88780 J/mole.

7 0
3 years ago
A solution is made by dissolving
nikklg [1K]

Answer:

THE MOLARITY IS 2.22 MOL/DM3

Explanation:

The solution formed was as a result of dissolving 37.5 g of Na2S in 217 g of water

Relative molecular mass of Na2S = ( 23* 2 + 32) = 78 g/mol

Molarity in g/dm3 is the amount of the substance dissolved in 1000 g or 1 L of the solvent. So we have;

37.5 g of Na2S = 217 g of water

( 37.5 * 1000 / 217 ) g = 1000 g of water

So, 172.81 g/dm3 of the solution

So therefore, molarity in mol/dm3 = mol in g/dm3 / molar mass

Molarity = 172.81 g/dm3 / 78 g/mol

Molarity = 2.22 mol/dm3

The molarity of the solution is 2.22 mol/dm3

4 0
3 years ago
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