What mass of potassium iodide is needed to prepare 5.60 L of a 1.13 M solution

1 answer:

8 0

Given that the question gives us concentration (M) and volume, we can use these so get moles. remember that Molarity (M)= mol/ Liters, so if we want the moles, then

moles= M x L

moles= 1.13 x 5.60= 6.33 moles KI

now to get the mass in grams, we need the molar mass of potassium iodide (KI) which can be determined using the periodic table and add the masses of each atom.

molar mass KI= 39.1 + 127= 166.1 g/mol

6.33 mol (166 g/ 1 mol)= 1050 grams KI

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The correct answer is because the molecular structure.

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3 years ago

It is advised that the bromobenzene solution be added slowly to the magnesium-ether solution so that it isn't present in a high

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Liquid Q is a polar solvent and liquid R is a nonpolar solvent. On the basis of this information, you would expect:

bearhunter [10]

4) is correct
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At 2000°C, the equilibrium constant for the reaction below is Kc = 4.10 ´ 10–4 . If 0.600 moles of NO is placed in a 1.0-L react

erastova [34]

Answer:

At equilibrium, the concentration of $N_{2 (g)}$ is going to be 0.30M

Explanation:

We first need the reaction.

With the information given we can assume that is:

$N_{2 (g)}$ + $O_{2 (g)}$ ⇄ 2 $NO_{(g)}$

If there is placed 0.600 moles of NO in a 1.0-L vessel, we have a initial concentration of 0.60 M NO; and no $N_{2 (g)}$ nor $O_{2 (g)}$ present. Immediately, $N_{2 (g)}$ and $O_{2 (g)}$ are going to be produced until equilibrium is reached.

By the ICE (initial, change, equilibrium) analysis:

I: [ $N_{2 (g)}$ ]=0 ; [ $O_{2 (g)}$ ]= 0 ; [ $NO_{(g)}$ ]=0.60M

C: [ $N_{2 (g)}$ ]=+x ; [ $O_{2 (g)}$ ]= +x ; [ $NO_{(g)}$ ]=-2x

E: [ $N_{2 (g)}$ ]=0+x ; [ $O_{2 (g)}$ ]= 0+x ; [ $NO_{(g)}$ ]=0.60-2x

Now we can use the constant information:

$K_{c}=\frac{[products]^{stoichiometric coefficient} }{[reactants]^{stoichiometric coefficient} }$