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Klio2033 [76]
3 years ago
15

The ending of a compounds name that is made up of three elements,one of which is oxygen​

Chemistry
1 answer:
Nezavi [6.7K]3 years ago
7 0

Answer:

—COOH

Explanation:

—COOH, acid group, containing 3 elements, C, carbon, O, oxygen, and H, hydrogen.

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3 years ago
An explanation for a broad of observation, facts, and tested hypothses is called a
nikitadnepr [17]

Answer:

that's is a theory

Explanation:

hope it may help you

4 0
2 years ago

grin007 [14]

Answer:

Calcium

Explanation:

The symbol for the element with a mass number of 27 is actually Al for Aluminum instead of Co for Cobalt. Mass number refers to atomic mass, not atomic number.

Aluminum has an atomic mass of 26.982 or 27.

The letters to unscramble are I, M, C, Al, U, and C.

The mystery element is Calcium.

Hope that helps.

3 0
3 years ago
At a certain temperature, the K p Kp for the decomposition of H 2 S H2S is 0.834 . 0.834. H 2 S ( g ) − ⇀ ↽ − H 2 ( g ) + S ( g
stira [4]

Answer:

Total pressure at equilibrium is 0.2798atm.

Explanation:

For the reaction:

H₂S(g) ⇄ H₂(g) + S(g)

Kp is defined as:

Kp = \frac{P_{H_{2}}*P_S}{P_{H_{2}S}} = 0.834

If initial pressure of H₂S is 0.150 atm, equilibrium pressures are:

H₂S(g): 0.150atm - x

H₂(g): x

S(g): x

Replacing in Kp:

\frac{X*X}{0.150atm-X} = 0.834

X² = 0.1251 - 0.834X

X² +  0.834X - 0.1251 = 0

Solving for X:

X = -0.964 → False solution: There is no negative pressures

X = 0.1298

Thus, pressures are:

H₂S(g): 0.150atm - 0.1298atm = <em>0.0202atm</em>

H₂(g): <em>0.1298atm</em>

S(g): <em>0.1298atm</em>

Thus, total pressure in the container at equilibrium is:

0.0202atm + 0.1298atm + 0.1298atm = <em>0.2798atm</em>

5 0
3 years ago
Read 2 more answers
A sample of oxygen gas was found to effuse at a rate equal to two times that of an unknown gas. The molecular weight of the unkn
Pie

Answer: 128 g/mol

Explanation:

Graham's law states that  the rate of effusion of a gas is inversely proportional to the square root of the molar mass of its particles.

Mathematically, that is:

\frac{Rate_1}{Rate_2}=\sqrt{\frac{MolarMass_2}{MolarMass_1} }

Since, you know the ratio of two rates and the molar mass of one gas, you can calculate the molar mass of the other gas.

The molar mass of the oxygen molecule, O₂ = 2×16.0g/mol = 32.0 g/mol.

In the coming equations, I will use 32 g/mol for simplicity of writing.

\frac{Rate_1}{Rate_2}=2 \\ \\ \\ MolarMass_1=32g/mol\\\\\\ 2=\sqrt{\frac{MolarMass_2}{32g/mol} } \\ \\ \\ 4=\frac{MolarMass_2}{32g/mol}\\ \\ \\ MolarMass_2=128g/mol

So, the molecular mass of the unnknown gas is 128 g/mol.

6 0
3 years ago
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