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3 years ago

The molal boiling-point and freezing-point constants are different for different solvents.

Chemistry

2 answers:

RoseWind [281]3 years ago

8 0

Answer:

I think the answer is true, sorry if I am wrong

Explanation:

Vadim26 [7]3 years ago

3 0

The right answer for this is false

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1. Mg2+ and Ca2+ are in the same group on the periodic table. In terms of electronic

finlep [7]

Answer:

Se detailed explanation.

Explanation:

Hello,

In this case, since both magnesium and calcium ions are in group IIA, we can review the following similar properties:

- Since both calcium and magnesium are in group IIA they have two valence electrons, it means that the both of them have two electrons at their outer shells.

- They are highly soluble in water when forming ionic bonds with nonmetals such as those belonging to halogens and oxygen's family.

- Calcium has 18 electrons and magnesium 10 which are two less than the total protons (20 and 12 respectively) since the both of them have lost two electrons due their ionized form.

- Their electron configurations are:

$Ca^{20}=1s^2,2s^2,2p^6,3s^3,3p^6,4s^2\\\\Mg^{12}=1s^2,2s^2,2p^6,3s^2$

It means that the both of them are at the $s$ region since it is the last subshell at which their electrons are.

Best regards.

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3 years ago

What is the evaluation of 60.1 x -9.2?

poizon [28]

That's the evaluation

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3 years ago

Help, please!

professor190 [17]

Hi so from what I can see the pizza as a distant or and you just have to convert the grams of glucose into moles. Most teachers ask for this format

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3 years ago

What happens when a parking lot is built in a wetlands area?

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The ground could sink and sink holes could occur, otherwise the parking lot could simply break apart and wear faster

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3 years ago

Read 2 more answers

Calculate the ph of a dilute solution that contains a molar ratio of potassium acetate to acetic acid (pka ???? 4.76) of (a) 2:1

givi [52]

According to Hasselbach-Henderson equation:

$pH=pK_{a}+log\frac{[A^{-}]}{[HA]}$

Here, $[A^{-}]$ is concentration of conjugate base and [HA] is concentration of acid.

In the given problem, conjugate base is $CH_{3}COOK$ and acid is $CH_{3}COOH$ thus, Hasselbach-Henderson equation will be as follows:

$pH=pK_{a}+log\frac{[CH_{3}COOK]}{[CH_{3}COOH]}$ ...... (1)

(a) Ratio of concentration of potassium acetate and acetic acid is 2:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=2$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{2}{1}=5.06$

Therefore, pH of solution is 5.06.

(b) Ratio of concentration of potassium acetate and acetic acid is 1:3 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{3}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{3}=4.28$

Therefore, pH of solution is 4.28.

(c)Ratio of concentration of potassium acetate and acetic acid is 5:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{5}{1}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{5}{1}=5.45$

Therefore, pH of solution is 5.45.

(d) Ratio of concentration of potassium acetate and acetic acid is 1:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=1$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{1}=4.76$

Therefore, pH of solution is 4.76.

(e) Ratio of concentration of potassium acetate and acetic acid is 1:10 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{10}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{10}=3.76$

Therefore, pH of solution is 3.76.

4 0

3 years ago