How many grams are in 27 L of O2 gas at STP

Rank the four gases (air, exhaled air, gas produced from the decomposition of H2O2, gas from decomposition of NaHCO3) in order o

Komok [63]

Astronomers use chemical signatures to determine the age and .... Gas mixtures that contain more than 4% hydrogen in air are potentially explosive. ... water vapor, carbon dioxide, and several other gases

6 0

3 years ago

The pOH of a solution is 4.2. Which of the following is true about the solution?

den301095 [7]

Answer:

It is basic and has a pH of 9.8.

Explanation:

pOH = 4.2

we will determine its pH.

pOH + pH = 14

pH = 14 - pOH

pH = 14 - 4.2

pH = 9.8

According to pH scale the the pH lower than 7 is consider acidic while pH of seven is neutral and pH greater than 7 is basic.

The given solution has pH 9.8 it means it is basic.

4 0

3 years ago

Why wouldn't two cations form a covalent bond?

Allushta [10]

Answer:

Covalent bonding is the sharing of electrons between atoms. In addition, the ionization energy of the atom is too large and the electron affinity of the atom is too small for ionic bonding to occur.

For example: carbon does not form ionic bonds because it has 4 valence electrons, half of an octet.

Explanation:

4 0

3 years ago

You are given a solid that is a mixture of Na2SO4 and K2SO4. A 0.205-g sample of the mixture is dissolved in water. An excess of

Svetach [21]

Answer:

Mass of SO₄⁻² = 0.123 g.

Mass percentage of SO₄⁻² = 41.2%

Mass of Na₂SO₄ = 0.0773 g

Mass of K₂SO₄ = 0.1277 g

Explanation:

Here we have

We place Na₂SO₄ = X and

K₂SO₄ = Y

Therefore

X +Y = 0.205 .........(1)

Therefore since the BaSO₄ is formed from BaCl₂, Na₂SO₄ and K₂SO₄ we have

Amount of BaSO₄ from Na₂SO₄ is therefore;

$X\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, Na_2SO_4}$

Amount of BaSO₄ from K₂SO₄ is;

$Y\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, K_2SO_4}$

Molar mass of

BaSO₄ = 233.38 g/mol

Na₂SO₄ = 142.04 g/mol

K₂SO₄ = 174.259 g/mol

$X\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, Na_2SO_4}$ = $X\times\frac{233.38 }{142.04}$ = 1.643·X

$Y\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, K_2SO_4}$ = $Y\times\frac{233.38 }{174.259 }$ = 1.339·Y

Therefore, we have

1.643·X + 1.339·Y = 0.298 g.....(2)

Solving equations (1) and (2) gives

The mass of SO₄⁻² in the sample is given by

Mass of sample = 0.298

Molar mass of BaSO₄ = 233.38 g/mol

Mass of Ba = 137.327 g/mol

∴ Mass of SO₄ = 233.38 g - 137.327 g = 96.05 g

Mass fraction of SO₄⁻² in BaSO₄ = 96.05 g/233.38 g = 0.412

Mass of SO₄⁻² in the sample is 0.412×0.298 = 0.123 g.

Percentage mass of SO₄⁻² = 41.2%

Solving equations (1) and (2) gives

X = 0.0773 g and Y = 0.1277 g.

8 0

3 years ago

How many moles of ammonium ions are in 1.3 grams of ammonium carbonate?

Eva8 [605]

Molecular weight of (NH4)2 CO3 = 96 gm. 96 gm (NH4)2CO3 = 1 mol of (NH4)2CO3 = 2 moles of NH4. Therefore 1.3 gm of (NH4)2CO3 = 1.3x2/ 96 = 0.027 moles.

3 0

3 years ago