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umka2103 [35]
3 years ago
8

What is the mass of 3.0 x 10^23 atoms of neon

Chemistry
1 answer:
anzhelika [568]3 years ago
8 0


3.0e23 atoms Ne

"E" means 10^

Then we multiply it by a mole of Ne. By the definetion of a mole, it is always 6.022e23 atoms of an element.

So now, we do this:

3.0e23 atoms Ne x (1 mol Ne / 6.022e23 atoms Ne)

After that, we use molar mass. A mole of Neon is equal, in terms of grams, to its avg. atomic mass. This goes true for any element.

It ends up like this:

3.0e23 atoms Ne x (1 mol Ne / 6.022e23 atoms Ne) x (20.1797 g Ne / 1 mol Ne)

Now cancel out the "atoms Ne" and "1 mol Ne"

You end up with a grand total of...
*plugs everything into a calculator*

10.05298... g Ne.

We need to round to 2 sig. figs. (3.0) so now it's....

10 g Ne.

Note that this method can only be used for converting atoms of an element to mass in grams.
Source(s):
A periodic table for the atomic mass of neon.
A chemistry textboook
A chemistry class.
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If a solution contains 3 moles/liter of sodium chloride (nacl, made of sodium ions and chloride ions), what is the osmolarity of
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<u>Answer:</u> Osmolarity of the sodium chloride solution is 18 Osmol/L.

<u>Explanation:</u>

Osmolarity is defined as the the concentration of the solution which is expressed as the total number of solution particles present in one liter of solvent.

We are given the molarity of the solution which is 3mol/L and to convert it into osmolarity, we will multiply the number of osmoles that are produced by the solute.

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5 0
3 years ago
The ph of a 0.0100 m solution of the sodium salt of a weak acid is 11.00. what is the ka of the acid?
Vitek1552 [10]
The answer is Ka = 1.00x10^-10.
Solution:
When given the pH value of the solution equal to 11, we can compute for pOH as
     pOH = 14 - pH = 14 - 11.00 = 3.00
We solve for the concentration of OH- using the equation
     [OH-] = 10^-pOH = 10^-3 = x

Considering the sodium salt NaA in water, we have the equation
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hence, [A-] = 0.0100 M

Since HA is a weak acid, then A- must be the conjugate base and we can set up an ICE table for the reaction
                             A- + H2O ⇌ HA + OH-
     Initial             0.0100            0       0
     Change        -x                    +x     +x
     Equilibrium    0.0100-x         x       x

We can now calculate the Kb for A-:
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Approximating that x is negligible compared to 0.0100 simplifies the equation to
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We can finally calculate the Ka for HA from the Kb, since we know that Kw = Ka*Kb = 1.0 x 10^-14:
     Ka = Kw / Kb 
           = 1.00x10^-14 / 1.00x10^-4
           = 1.00x10^-10
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