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3 years ago

What is the element found in Group 17 Period 4?

Chemistry

1 answer:

faust18 [17]3 years ago

3 0

Answer:

Bromine is found in Group 17, Period 4 on the periodic table.

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Given the balanced equation below, how many moles of chromium are

fenix001 [56]

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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

What is 11.75 millimeters in meters

Lana71 [14]

Answer:

in meters its 0.01175

Explanation:

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1 year ago

Read 2 more answers

Biphenyl and benzene are common side products in this gringard reaction. explain at which points in the purification procedure t

mafiozo [28]

In Grignard reaction, Biphenyl and benzene are common side products which are removed during trituration.

In organic chemistry, a reaction in which the Grignard reagents or organometallic substances are added to organic compounds such as aldehydes and ketones to form alcohol is known as Grignard reaction.

These Grignard reagents are magnesium halides of alkyl, vinyl or allyl, which react with a carbonyl group to form alcohols.

During this reaction, primary, secondary and tertiary alcohols are formed.

While Biphenyl and benzene are common side products.

These are removed during trituration process in which cold petroleum ether is added to dissolve the biphenyl and benzene side products

If you need to learn more about Biphenyl and benzene click here:

brainly.com/question/4336669

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1 year ago

WHY IS IT IMPORTANT TO HAVE A LARGE SAMPLE SIZE IN ANY EXPERIMENT?

Anon25 [30]

It is important because if the sample size is smaller, outliers could skew the data more than if it was large.

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3 years ago

What is the element found in Group 17 Period 4?​

What is the element found in Group 17 Period 4?