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BigorU [14]
3 years ago
13

An empty 10-ml volumetric flask weighs 10.267 g. after filling to the mark with distilled water at 20°c, the mass is 20.219 g. w

hat is the mass of water in the flask
Chemistry
1 answer:
Dahasolnce [82]3 years ago
3 0
Ok, so... 20.219 grams is the mass of the flask with the water in it.
10.267 grams is the mass of only the flask. Subtract them to find the water mass.

20.219 g (flask+water) - 10.267 g (flask) = 9.952 g (water)
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marta [7]

Molality of the solution is defined as the number of moles of a substance dissolved divided by the mass of the solvent:

Molality = number of moles / solvent mass

From the concentration of 39% (by mass) of HCl in water, we construct the following  reasoning:

in 100 g solution we have 39 g hydrochloric acid (HCl)

number of moles  = mass / molecular weight

number of moles of HCl = 39 / 36.5 = 1.07 moles

solvent (water) mass = solution mass - hydrochloric acid mass

solvent (water) mass = 100 - 39 = 61 g

Now we can determine the molality:

molality = 1.07 moles / 61 g = 0.018

8 0
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What does it mean for the energy levels in an atom to be quantized? Describe how this affects where electrons can be found
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3 0
3 years ago
A galvanic cell consists of one half-cell that contains Ag(s) and Ag+(aq), and one half-cell that contains Cu(s) and Cu2+(aq). W
Agata [3.3K]

Answer : The 'Ag' is produced at the cathode electrode and 'Cu' is produced at anode electrode under standard conditions.

Explanation :

Galvanic cell : It is defined as a device which is used for the conversion of the chemical energy produces in a redox reaction into the electrical energy. It is also known as the voltaic cell or electrochemical cell.

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We are taking the value of standard reduction potential form the standard table.

E^0_{[Ag^{+}/Ag]}=+0.80V

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In this cell, the component that has lower standard reduction potential gets oxidized and that is added to the anode electrode. The second forms the cathode electrode.

The balanced two-half reactions will be,

Oxidation half reaction (Anode) : Cu(s)\rightarrow Cu^{2+}(aq)+2e^-

Reduction half reaction (Cathode) : Ag^{+}(aq)+e^-\rightarrow Ag(s)

Thus the overall reaction will be,

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

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Hence, the 'Ag' is produced at the cathode electrode and 'Cu' is produced at anode electrode under standard conditions.

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