All categories

2 years ago

Join anyone qof-gsqi-tnk

Chemistry

2 answers:

vampirchik [111]2 years ago

7 0

i don know sowrry:) hehe

dem82 [27]2 years ago

4 0

Answer:

$not \: intrested \: \\ thnx \: for \: \\ pts$

You might be interested in

If you had 100 ml of juice, how many milliliter would de fruit juice?

Julli [10]

Since the measurement is not changing, the answer is 100 mL. Hope this helps.

8 0

3 years ago

The reaction for the formation of gaseous hydrogen fluoride (HF) from molecular hydrogen (H2) and fluorine (F2) has an equilibri

mestny [16]

Answer:

[H2]eq = 0.0129 M

[F2]eq = 1.0129 M

[HF]eq = 0.9871 M

Explanation:

H2(g) + F2(g) ↔ 2HF(g)

∴ Ke = [HF]² / [H2]*[F2] = 1.15 E2

experiment:

∴ n H2 = 3.00 mol

∴ n F2 = 6.00 mol

∴ V sln = 3.00 L

⇒ [H2]i = 3.00 mol / 3.00 L = 1 M

⇒ [F2]i = 6.00 mol / 3.00 L = 2 M

[ ]i change [ ]eq

H2 1 1 - x 1 - x

F2 2 2 - x 2 - x

HF - x x

⇒ K = (x)² / (1 - x)*(2 - x) = 1.15 E2

⇒ x² / (2 - 3x + x²) = 1.15 E2 = 115

⇒ x² = (2 - 3x + x²)(115)

⇒ x² = 230 - 345x + 115x²

⇒ 0 = 230 - 345x + 114x²

⇒ x = 0.9871

equilibrium:

⇒ [H2] = 1 - x = 1 - 0.9871 = 0.0129 M

⇒ [F2] = 2 - x = 2 - 0.9871 = 1.0129 M

⇒ [HF] = x = 0.9871 M

5 0

3 years ago

A 17g sample of H2O2 was decomposed to yield 1g of H2 and 16g of O2 An unknown sample containing only H and O was decomposed to

ludmilkaskok [199]

Answer:

Explanation:

Our $H_2O_2$ sample yielded 1g of $H_2$ and 16g of $O_2$ , but our unknown sample yielded 2 times as much $H_2$ for the same amount of $O_2$ .

What does this mean? that the H:O proportion for the unknown sample is twice the H:O proportion for the $H_2O_2$ sample.

What is the H:O proportion for the $H_2O_2$ sample? As we can see from its formula, it's 1:1, therefore the proportion for the unknown formula must be 2:1.

That means, two H atoms for every O atom. We could write that as: $H_2O$ and you should recognize that formula, for it is one of the most common compounds on earth, Water.

5 0

3 years ago

A buffer is prepared by adding 300. 0 ml of 2. 0 mnaoh to 500. 0 ml of 2. 0 mch3cooh. what is the ph of this buffer? ka= 1. 8 10

Anton [14]

The Henderson-Hasselbalch equation can be used to determine the pH of the buffer from the pKa value. The pH of the buffer will be 4.75.

<h3>What is the Henderson-Hasselbalch equation?</h3>

Henderson-Hasselbalch equation is used to determine the value of pH of the buffer with the help of the acid disassociation constant.

Given,

Acid disassociation constant (ka) = 1. 8 10⁻⁵

Concentration of NaOH = 2.0 M

Concentration of CH₃COOH = 2.0 M

pKa value is calculated as,

pKa = -log Ka

pKa = - log (1. 8 x 10⁻⁵)

Substituting the value of pKa in the Henderson-Hasselbalch equation as

pH = - log (1. 8 x 10⁻⁵) + log [2.0] ÷ [2.0]

pH = - log (1. 8 x 10⁻⁵) + log [1]

= 4.745 + 0

= 4.75

Therefore, 4.75 is the pH of the buffer.

Learn more about the Henderson-Hasselbalch equation here:

brainly.com/question/27751586

#SPJ4

6 0

1 year ago

Compare the reactivity between lithium and sodium towards oxygen.

Scrat [10]

Reactivity of Group 1 and 2 elements increases as you go down the periodic table. So sodium is more reactive than lithium. Sodium will react with oxygen forming Na2O (sodium oxide). Lithium forms lithium oxide (Li2O).

7 0

2 years ago

Join anyone qof-gsqi-tnk​

Join anyone qof-gsqi-tnk