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zhuklara [117]
3 years ago
6

8. An experiment requires a solution that is 80%

Chemistry
1 answer:
konstantin123 [22]3 years ago
4 0

Answer:

v_{solute}=160mL

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the volume of methyl alcohol solute by using the definition of by-volume percentage:

\%v=\frac{v_{solute}}{v_{solution}} *100\%

Whereas we solve for the volume of the solute as shown below:

v_{solute}=\frac{\%v*v_{solution}}{100\%} \\\\v_{solute}=\frac{80\%*200mL}{100\%}\\\\ v_{solute}=160mL

Regards!

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Answer : The value of equilibrium constant (Kc) is, 0.0154

Explanation :

The given chemical reaction is:

                        SO_2Cl_2(g)\rightarrow SO_2(g)+Cl_2(g)

Initial conc.    2.4\times 10^{-2}          0             0

At eqm.          (2.4\times 10^{-2}-x)   x              x

As we are given:

Concentration of Cl_2 at equilibrium = 1.3\times 10^{-2}M

That means,

x=1.3\times 10^{-2}M

The expression for equilibrium constant is:

K_c=\frac{[SO_2][Cl_2]}{[SO_2Cl_2]}

Now put all the given values in this expression, we get:

K_c=\frac{(x)\times (x)}{2.4\times 10^{-2}-x}

K_c=\frac{(1.3\times 10^{-2})\times (1.3\times 10^{-2})}{2.4\times 10^{-2}-1.3\times 10^{-2}}

K_c=0.0154

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3 years ago
How many molecules is 7.22 mole H2O
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Explanation:

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Which of the following is true for the reaction below? CO2(g) + 393.5 kJ C(s) + O2(g)
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It is endothermic and the ∆H = +393.5 kJ.
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What is the empirical formula of a compound that is 24.42 % calcium, 17.07 % nitrogen, and 58.5% oxygen?
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CaN_{2} O_{6}

Explanation:

When calculating an empirical formula from percentages, assume you have a 100g sample. This allows you to convert the percentages directly to grams, because X % of 100g is X grams.

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The next step is to divide each mass by their molar mass to convert your grams to moles.

24.42/40.08 = 0.6092 mol

17.07/14.01 = 1.218 mol

58.85/15.99 = 3.680 mol

Then you will divide all of your mol values by the SMALLEST number of moles. This gives you whole numbers that are the mole ratio (subcripts) of the empircal formula.

0.6092 mol/0.6092 mol = 1

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