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 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
NaA → Na+ + A-
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-:
Kb = [HA][OH-] / [A-]
= x<span>²</span> / 0.0100-x
Approximating that x is negligible compared to 0.0100 simplifies the equation to
Kb = (10^-3)² / 0.0100 = 0.000100 = 1.00x10^-4

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
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3 years ago

Consider three gases: Ar, SF6, and Cl2. If 50.0 grams of these gases are placed in each of three identical containers, which con

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The ideal gas law:
$pV=nRT \Rightarrow p=\frac{nRT}{V}$
p - pressure, n - number of moles, R - the gas constant, T - temperature, V - volume

The volume and temperature of all three containers are the same, so the pressure depends on the number of moles. The greater the number of moles, the higher the pressure.
The mass of gases is 50 g.

$Ar \\
M \approx 39.948 \ \frac{g}{mol} \\
n=\frac{50 \ g}{39.948 \ \frac{g}{mol}} \approx 1.25 \ mol \\ \\
SF_6 \\
M \approx 146.06 \ \frac{g}{mol} \\
n=\frac{50 \ g}{146.06 \ \frac{g}{mol}} \approx 0.34 \ mol \\ \\
Cl_2 \\
M=70.9 \ \frac{g}{mol} \\
n=\frac{50 \ g}{70.9 \ \frac{g}{mol}} \approx 0.71 \ mol$

The greatest number of moles is in the container with Ar, so there is the highest pressure.

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3 years ago

What is the molarity of a solution that contains 11.6 moles of LiNO3 in 13.7 liters of solution?

Tatiana [17]

The molarity of this should be around 0.846

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Formula: M= mols/liters

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So… M= 11.6/13.7⇒ 0.846

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2 years ago