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Aleks04 [339]
3 years ago
12

Which contains more molecules; a mole of Chlorine gas (Cl2 ) or a mole of glucose (C6H1206) ?

Chemistry
1 answer:
Elodia [21]3 years ago
7 0
This is a tricky question. a mole of any compound contains the same number of molecules of that certain compound. so, one mole of chlorine gas has the same number of molecules as one mole of glucose, which is 6.02 x 10^23.

this is avogadro's number and it applies for any mole of molecules. 

the question is tricky because it is like asking. " what weighs more, a pound of feathers or a pound of rocks?" the both weigh the same, a pound. when ewe talking about moles, same as pounds, it is a quantity unit. one mole will aways be equal to 6.02 x 10^23 molecules.
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How many formula units of silver fluoride, AgF, are equal to 42.15 g f this substance?
s2008m [1.1K]

Answer:

1.99 x 10²³ formula units

Explanation:

Given parameters:

Mass of AgF  = 42.15g

Unknown:

The amount of formula units

Solution:

To solve this problem, we set out by find the number of moles in this compound from the given mass.

   Number of moles  = \frac{mass}{molar mass}

molar mass of AgF  = 107.9 + 19  = 126.9g/mol

  Number of moles  = \frac{42.15}{126.9}   = 0.33 moles

       1 mole of a substance =  6.02 x 10²³ formula units

         0.33 moles of AgF  = 0.33 x 6.02 x 10²³  = 1.99 x 10²³ formula units

7 0
3 years ago
A student does an experiment to determine the molar solubility of lead(II) bromide. She constructs a voltaic cell at 298 K consi
attashe74 [19]

Answer:

The molar solubility of lead bromide at 298K is 0.010 mol/L.

Explanation:

In order to solve this problem, we need to use the Nernst Equaiton:

E = E^{o} - \frac{0.0591}{n} log\frac{[ox]}{[red]}

E is the cell potential at a certain instant, E⁰ is the cell potential, n is the number of electrons involved in the redox reaction, [ox] is the concentration of the oxidated specie and [red] is the concentration of the reduced specie.

At equilibrium, E = 0, therefore:

E^{o}  = \frac{0.0591}{n} log \frac{[ox]}{[red]} \\\\log \frac{[ox]}{[red]} = \frac{nE^{o} }{0.0591} \\\\log[red] =  log[ox] -  \frac{nE^{o} }{0.0591}\\\\[red] = 10^{ log[ox] -  \frac{nE^{o} }{0.0591}} \\\\[red] = 10^{ log0.733 -  \frac{2x5.45x10^{-2}  }{0.0591}}\\\\

[red] = 0.010 M

The reduction will happen in the anode, therefore, the concentration of the reduced specie is equivalent to the molar solubility of lead bromide.

7 0
3 years ago
How close does Venus get to earth
Sergio [31]
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How many grams of NaOH are produced from 20 grams of Na2Co3
Mariulka [41]

Answer:

1400

Explanation:

4 0
3 years ago
2HCl+Ca(OH)2⟶CaCl2+2H2O
RUDIKE [14]
Isn’t it just 3 mols?
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