What use is chlorine-36

You are given an unknown gaseous binary compound (that is, a compound consisting of two different elements). when 10.0 g of the

Luda [366]

Answer is: a possible identity for the unknown compound is C₃H₈.

m(binary compound) = 10.0 g.

m(H₂O) = 16.3 g; mass of water.

M(O₂) = 32 g/mol; molar mass of oxygen.

M(binary compound) = 1.38 · 32 g/mol.

M(binary compound) = 44.16 g/mol.

n(binary compound) = 10 g ÷ 44.16 g/mol.

n(binary compound) = 0.225 mol; amount of substance.

n(H₂O) = 16.3 g ÷ 18 g/mol.

n(H₂O) = 0.9 mol; amount of water.

m(H₂O) : n(binary compound) = 0.9 mol ÷ 0.225 mol.

m(H₂O) : n(binary compound) = 4 : 1.

Unknown cpmpound has 4 times more hydrogen than water, it has 8 hydrogen atoms.

Second element in compound is carbon:

M(X) = 44.16 g/mol - 8 · 1.01 g/mol.

M(X) = 36.08 g/mol ÷ 3.

M(C) = 12.01 g/mol.

8 0

3 years ago

In the EXPLORE section of your lesson 4.08 on Potential energy there were several animations to watch that provided a graphic il

Lunna [17]

Answer:

This is because no energy is being created or destroyed in this system

Explanation:

I think this is correct? I hope it helps.

7 0

3 years ago

Read 2 more answers

What is the mass of oxygen in 3.34 g of potassium permanganate?

3241004551 [841]

Answer:

1.35 g

Explanation:

Data Given:

mass of Potassium Permagnate (KMnO₄) = 3.34 g

Mass of Oxygen: ?

Solution:

First find the percentage composition of Oxygen in Potassium Permagnate (KMnO₄)

So,

Molar Mass of KMnO₄ = 39 + 55 + 4(16)

Molar Mass of KMnO₄ = 158 g/mol

Calculate the mole percent composition of Oxygen in Potassium Permagnate (KMnO₄).

Mass contributed by Oxygen (O) = 4 (16) = 64 g

Since the percentage of compound is 100

So,

Percent of Oxygen (O) = 64 / 158 x 100

Percent of Oxygen (O) = 40.5 %

It means that for ever gram of Potassium Permagnate (KMnO₄) there is 0.405 g of Oxygen (O) is present.

So,

for the 3.34 grams of Potassium Permagnate (KMnO₄) the mass of Oxygen will be

mass of Oxygen (O) = 0.405 x 3.34 g

mass of Oxygen (O) = 1.35 g

5 0

3 years ago

A solution of NaF is added dropwise to a solution that is 0.0144 M in Ba 2 . When the concentration of F - exceeds __________ M,

Stella [2.4K]

Answer:

When [F⁻] exceeds 0.0109M concentration, BaF₂ will precipitate

Explanation:

Ksp of BaF₂ is:

BaF₂(s) ⇄ Ba²⁺(aq) + 2F⁻(aq)

Ksp = 1.7x10⁻⁶ = [Ba²⁺] [F⁻]²

The solution will produce BaF₂(s) -precipitate- just when [Ba²⁺] [F⁻]² > 1.7x10⁻⁶.

As the concentration of [Ba²⁺] is 0.0144M, the product [Ba²⁺] [F⁻]² will be equal to ksp just when:

1.7x10⁻⁶ = [Ba²⁺] [F⁻]²

1.7x10⁻⁶ = [0.0144M] [F⁻]²

1.18x10⁻⁴ = [F⁻]²

0.0109M = [F⁻]

That means, when [F⁻] exceeds 0.0109M concentration, BaF₂ will precipitate

5 0

3 years ago

Match each substance with the correct designation for the equation HSO3- + CH3NH2 <=> SO32- + CH3NH3+ HSO3- CH3NH2 SO32- C

Zanzabum

Answer:

$HSO_3^-$ : conjugate acid of $SO_3^{2-}$

$CH_3NH_2$ : conjugate base of $CH_3NH_3^+$

$SO_3^{2-}$ : conjugate base of $HSO_3^-$

$CH_3NH_3^+$ : conjugate acid of $CH_3NH_2$

Explanation:

According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.

$HSO_3^-+CH_3NH_2\rightleftharpoons SO_3^{2-}+CH_3NH_3^+$

Here in forward reaction $CH_3NH_2$ is accepting a proton, thus it is considered as a base and after accepting a proton, it forms $CH_3NH_3^+$ which is a conjugate acid.

And $HSO_3^-$ is losing a proton, thus it is considered as an acid and after loosing a proton, it forms $SO_3^{2-}$ which is a conjugate base.

Similarly in the backward reaction, $CH_3NH_3^+$ is loosing a proton, thus it is considered as a acid and after loosing a proton, it forms $CH_3NH_2$ which is a conjugate base.

And $SO_3^{2-}$ is accepting a proton, thus it is considered as a base and after accepting a proton, it forms $HSO_3^{-}$ which is a conjugate acid.

4 0

3 years ago