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3 years ago

Convert 5.0x10^24 molecules to liters.

1 answer:

3 0

To convert the given value we need conversion factors to relate molecules to liters. At STP, we know that 1 mol is equal to 22.4 L and by using Avogadro's number we can relate molecules to 1 mol. Calculation is as follows:

5.0x10^24 molecules ( 1 mol / 6.022 x 10^23 molecules ) ( 22.4 L / 1 mol) = 186.0 L

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150 mL of 0.25 mol/L magnesium chloride solution and 150 mL of 0.35 mol/L silver nitrate solution are mixed together. After reac

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Answer:

$0.175\; \rm mol \cdot L^{-1}$ .

Explanation:

Magnesium chloride and silver nitrate reacts at a $2:1$ ratio:

$\rm MgCl_2\, (aq) + 2\, AgNO_3\, (aq) \to Mg(NO_3)_2 \, (aq) + 2\, AgCl\, (s)$ .

In reality, the nitrate ion from silver nitrate did not take part in this reaction at all. Consider the ionic equation for this very reaction:

$\begin{aligned}& \rm Mg^{2+} + 2\, Cl^{-} + 2\, Ag^{+} + 2\, {NO_3}^{-} \\&\to \rm Mg^{2+} + 2\, {NO_3}^{-} + 2\, AgCl\, (s)\end{aligned}$ .

The precipitate silver chloride $\rm AgCl$ is insoluble in water and barely ionizes. Hence, $\rm AgCl\!$ isn't rewritten as ions.

Net ionic equation:

$\begin{aligned}& \rm Ag^{+} + Cl^{-} \to AgCl\, (s)\end{aligned}$ .

Calculate the initial quantity of nitrate ions in the mixture.

$\begin{aligned}n(\text{initial}) &= c(\text{initial}) \cdot V(\text{initial}) \\ &= 0.25\; \rm mol \cdot L^{-1} \times 0.150\; \rm L \\ &= 0.0375\; \rm mol \end{aligned}$ .

Since nitrate ions $\rm {NO_3}^{-}$ do not take part in any reaction in this mixture, the quantity of this ion would stay the same.

$n(\text{final}) = n(\text{initial}) = 0.0375\; \rm mol$ .

However, the volume of the new solution is twice that of the original nitrate solution. Hence, the concentration of nitrate ions in the new solution would be $(1/2)$ of the concentration in the original solution.

$\begin{aligned} c(\text{final}) &= \frac{n(\text{final})}{V(\text{final})} \\ &= \frac{0.0375\; \rm mol}{0.300\; \rm L} = 0.175\; \rm mol \cdot L^{-1}\end{aligned}$ .

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What would be the best indicator for titrating the weak base sodium bicarbonate using hcl titrant?

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The methyl orange would be the best indicator for titrating the weak base sodium bicarbonate using HCl titrant

When weak base is titrated with strong acid then , then solution is slightly acidic at end point . If weak acid is titrated with strong base then , the solution is slightly basic because salt formed will hydrolyzed to a certain extent .

In acid base titration at the end point the amount of the acid becomes chemically equivalent to the amount of the base present .The methyl orange would be the best indicator for titrating the weak base with strong acid .

To learn more about weak base please click here ,

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