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2 years ago

15

What kind of intermolecular force is KI(aq)

1 answer:

NeTakaya2 years ago

7 0

Answer:

dipole-induced dipole forces

Explanation:

KI(aq) has potassium and iodine ions in water. Water is a polar molecule. So the type of force must be ion-dipole.

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Briefly explain formality in solution.

Zielflug [23.3K]

Answer:

Both molarity and formality express concentration as moles of solute per liter of solution. Formality is a substance's total concentration in solution without regard to its specific chemical form. ... The formality of a solution is defined as the number of formula mass of any solute dissolved in 1 litre of solution.

4 0

2 years ago

A(n) _____ is a region where fresh water and salt water mix.

Airida [17]

It is called an estuary.

7 0

3 years ago

Start with 100.00 mL of 0.10 M acetic acid, CH3COOH. The solution has a pH of 2.87 at 25 oC. a) Calculate the Ka of acetic acid

jasenka [17]

Answer: a) The $K_a$ of acetic acid at $25^0C$ is $1.82\times 10^{-5}$

b) The percent dissociation for the solution is $4.27\times 10^{-3}$

Explanation:

$CH_3COOH\rightarrow CH_3COO^-H^+$

cM 0 0

$c-c\alpha$ $c\alpha$ $c\alpha$

So dissociation constant will be:

$K_a=\frac{(c\alpha)^{2}}{c-c\alpha}$

Give c= 0.10 M and $\alpha$ = ?

Also $pH=-log[H^+]$

$2.87=-log[H^+]$

$[H^+]=1.35\times 10^{-3}M$

$[CH_3COO^-]=1.35\times 10^{-3}M$

$[CH_3COOH]=(0.10M-1.35\times 10^{-3}=0.09806M$

Putting in the values we get:

$K_a=\frac{(1.35\times 10^{-3})^2}{(0.09806)}$

$K_a=1.82\times 10^{-5}$

b) $\alpha=\sqrt\frac{K_a}{c}$

$\alpha=\sqrt\frac{1.82\times 10^{-5}}{0.10}$

$\alpha=4.27\times 10^{-5}$

$\% \alpha=4.27\times 10^{-5}\times 100=4.27\times 10^{-3}$

5 0

3 years ago

What is the mass of 8.90 moles of magnesium chloride, MgCl2?

Korolek [52]

No of moles of MgCl2 = weight of MgCl2 / Molecular weight of MgCl2 Weight of MgCl2 =moles of MgCl2 x molecular mass of MgCl2 = 8.90 x 95=845.5 gm

6 0

3 years ago

A geological sample is found to have a Pb-206/U-238 mass ratio of 0.337/1.00. Assuming there was no Pb-206 present when the samp

azamat

Answer:

$Age=2.52*10^9 years$

Explanation:

To determine the age, we need to know how much U have become Pb, so first we have to calculate the Pb moles present in a sample:

$n_{Pb}=\frac{0.337g}{206g/mol}=0.00164 mol$

$n_{U}=\frac{1 g}{238g/mol}=0.0042 mol$

The percentage of U degradation:

$P=\frac{n_{Pb}}{n_{Pb}+n_{U}}$

$P=\frac{0.00164}{0.00164+0.0042}=0.28$

Assuming that the life time is linear:

$Age=\frac{4.5*10^9 years}{0.5 life time}*0.28 life time$

$Age=2.52*10^9 years$

4 0

3 years ago

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