Ask question

Ask question

All categories

Dmitry_Shevchenko [17]

3 years ago

6

What type of chemical bond is hydrogen?

1 answer:

oksian1 [2.3K]3 years ago

7 0

Hydrogen is a covalent bond. (A bond where one or more pairs of electrons are shared by two atoms)

You might be interested in

Draw the products formed when cholesterol is treated with [1] bh3·thf; [2] h2o2, −oh. indicate the stereochemistry around any st

Andreas93 [3]

Alkene on Hydration yield Alcohols. When Asymmetric Alkenes are treated with water in the presence of acid they follow Markovnikoff rule, and the Hydrogen of incoming reagent goes to that carbon which contain more number of Hydrogen atoms. The reverse (Anti-Markovnikoff) is done by carrying out Hydroboration reaction.In this case the Hydrogen atom of incoming reagent goes to that carbon which contains less number of Hydrogen atoms. In this reaction both -H and -OH adds in syn fashion. The Hydroboration reaction of Chloestrol is shown below,

3 0

3 years ago

What are the products in a neutralization reaction A. Two elements B.an acid and a base C. An element and a compound D. A salt a

frez [133]

D. Salt and water

Explanation:

Acid + Alkali -> Salt + water

5 0

3 years ago

Read 2 more answers

5. Which of the following statements is incorrect?

Vikki [24]

Answer:

c- dry cells can not be recharged

6 0

3 years ago

A cubic meter is about the same as the volume occupied by a _____. basketball arena kilogram of water washing machine cup of mil

Zinaida [17]

washing machine, hopefully this helps

6 0

3 years ago

What volumes of 0.200 M HCOOH and 2.00 M NaOH would make 500. mL of a buffer with the same pH as a buffer made from 475 mL of 0.

Hitman42 [59]

Explanation:

The given data is as follows.

[HCOOH] = 0.2 M, [NaOH] = 2.0 M,

V = 500 ml, [Benzoic acid] = 0.2 M

First, we will calculate the number of moles of benzoic acid as follows.

No. of moles of benzoic acid = Molarity × Volume

= $2 \times 0.475$

= 0.095 mol

And, moles of NaOH present in the solution will be as follows.

No. of moles of NaOH = Molarity × Volume

= $2 \times 0.025$

= 0.05 mol

Hence, the ICE table for the chemical equation will be as follows.

$C_{6}H_{5}COOH + NaOH \rightarrow C_{6}H_{5}COONa + H_{2}O$

Initial: 0.095 0.05 0 0

Equlbm: (0.095 - 0.05) 0 0.05

pH = $pK_{a} + log \frac{Base}{Acid}$

= $4.2 + log \frac{0.05}{0.045}$

= 4.245

For,

$HCOOH + NaOH \rightarrow HCOONa + H_{2}O$

Initial: 0.2x 2(0.5 - x) 0

Equlbm: 0.2x - 2(0.5 - x) 0 2(0.5 - x)

As,

pH = $pK_{a} + log \frac{Base}{Acid}$

4.245 = 3.75 + $log \frac{Base}{Acid}$

$log \frac{Base}{Acid}$ = 0.5

$\frac{Base}{Acid}$ = 3.162

Now,

$\frac{2(0.5 - x)}{0.2x - 2(0.5 - x)}$ = 3.162

x = 0.464 L

Volume of NaOH = (0.5 - 0.464) L

= 0.036 L

= 36 ml (as 1 L = 1000 mL)

And, volume of formic acid is 464 mL.

8 0

3 years ago