CrO and Cr₂O₃ make up the simplest chromium oxide formula.
What name does Cr₂O₃ use?
- Chromium oxide (Cr₂O₃)sometimes referred to as chromium sesquioxide or chromic oxide, is a compound in which chromium is oxidized to a +3 state. Sodium dichromate is calcined with either carbon or sulfur to produce it.
- Eskolaite, a mineral that bears the name of the Finnish geologist Pentti Eskola, is a kind of chromium oxide green that may be found in nature. The metallic glassy green surface of this unusual material has an unsettling moss-like look that may be used to conceal oneself in the environment.
- Studies on humans have conclusively shown that chromium (VI) breathed is a potential carcinogen, increasing the likelihood of developing lung cancer. According to animal studies, chromium (VI) exposure by inhalation can result in lung cancers.
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Answer:
Oxide of M is
and sulfate of 
Explanation:
0.303 L of molecular hydrogen gas measured at 17°C and 741 mmHg.
Let moles of hydrogen gas be n.
Temperature of the gas ,T= 17°C =290 K
Pressure of the gas ,P= 741 mmHg= 0.9633 atm
Volume occupied by gas , V = 0.303 L
Using an ideal gas equation:


Moles of hydrogen gas produced = 0.01225 mol

Moles of metal =
So, 8.3333 mol of metal M gives 0.01225 mol of hydrogen gas.

x = 2.9 ≈ 3


Formulas for the oxide and sulfate of M will be:
Oxide of M is
and sulfate of
.
Its known as covalently bonded atoms
Well, first we must remember that

This is because


So then

The molar mass of the unknown gas is 184.96 g/mol
<h3>Graham's law of diffusion </h3>
This states that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass i.e
R ∝ 1/ √M
R₁/R₂ = √(M₂/M₁)
<h3>How to determine the molar mass of the unknown gas </h3>
The following data were obtained from the question:
- Rate of unknown gas (R₁) = R
- Rate of CH₄ (R₂) = 3.4R
- Molar mass of CH₄ (M₂) = 16 g/mol
- Molar mass of unknown gas (M₁) =?
The molar mass of the unknown gas can be obtained as follow:
R₁/R₂ = √(M₂/M₁)
R / 3.4R = √(16 / M₁)
1 / 3.4 = √(16 / M₁)
Square both side
(1 / 3.4)² = 16 / M₁
Cross multiply
(1 / 3.4)² × M₁ = 16
Divide both side by (1 / 3.4)²
M₁ = 16 / (1 / 3.4)²
M₁ = 184.96 g/mol
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