What is the molarity when water is added to 4 moles of sodium chloride to make 0.5 liter of solution?

1 answer:

7 0

If 0.5 L of solution contains 4 mol
then let 1 L of solution contain x mol

⇒ (0.5 L) x = (4 mol) (1 L)

x = (4 mol · L) ÷ (0.5 L)

x = 8 mol

Thus the molarity of the Sodium Chloride solution is 8 mol / L OR 8 mol/dm³.

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What is the correct equilibrium constant expression for equation P2(g)

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Answer:

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Answer:

a. 1810mL

Explanation:

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To convert from Celsius to Kelvin, add 273, or use the equation: $T_C+273=T_K$

For this problem, one must also recall that standard temperature is 0°C (or 273K).

So, $T_1 = 273[K]$ , and $T_2 = (49.4+273)[K]=322.4[K]$ .

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$\dfrac{(1532.7[mL])}{(273[K\!\!\!\!\!{-}])}(322.4[K\!\!\!\!\!{-}] )=\dfrac{V_2}{(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})}(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})$