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3 years ago

Could someone help with this? Much appreciated!

Chemistry

1 answer:

alekssr [168]3 years ago

6 0

Answer:

The 3rd answer down.

Na²O (sodium oxide) will be a base when exposed to water H²O

Explanation:

Sodium Oxide Na²O, will become Sodium Hydroxide after being exposed to water (at 80% I believe).

The oxygen ion in Na²O has 2 extra electrons which makes it highly charged and very attractive to hydrogen ions. The attraction is so strong that when Na²O comes in contact with H²O, the O(-2) strips off a hydrogen from water, forming 2 x OH ions which of course are still strongly basic.

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Can someone help me with this

Flura [38]

Answer:

I don't know how to get the answer sorry

Explanation:

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2 years ago

Is methylene chloride more or less dense than water?

horsena [70]

Methylene chloride is less dense than water

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3 years ago

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.

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3 years ago

Help with 6-10 you will be marked brain list

Fofino [41]

Answer:

wanna waist my points so here

Explanation:

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3 years ago

Read 2 more answers

A chemistry student needs of carbon tetrachloride for an experiment. By consulting the CRC Handbook of Chemistry and Physics, th

mixer [17]

Answer:

34.6 cm³

Explanation:

<em>A chemistry student needs 55.0 g of carbon tetrachloride for an experiment. By consulting the CRC Handbook of Chemistry and Physics, the student discovers that the density of carbon tetrachloride is 1.59 g/cm³. Calculate the volume of carbon tetrachloride the student should pour out. Be sure your answer has the correct number of significant digits.</em>

Step 1: Given data

Mass of carbon tetrachloride (m): 55.0 g

Density of carbon tetrachloride (ρ): 1.59 g/cm³

Step 2: Calculate the required volume of carbon tetrachloride

Density is an intrinsic property of matter. It can be calculated as the quotient between the mass of the sample and its volume.

ρ = m/V

V = m/ρ

V = 55.0 g/(1.59 g/cm³)

V = 34.6 cm³

The chemistry student should pour 34.6 cm³ of carbon tetrachloride.

8 0

3 years ago