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2 years ago

In the compound AbO3, the ratio of aluminum to oxygen is

Chemistry

1 answer:

jasenka [17]2 years ago

3 0

The compound is Al2O3. The ratio of aluminum to oxygen is 2:3.

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1. An atom that has an outer shell that is NOT full is (more likely OR less likely) to be reactive than an atom that has a full

Serga [27]

Yes because they atom with a stable octet electrons is less reactive it does not need any more electrons while an atom that has an partially filled electrons need more electrons to complete octet structure

7 0

2 years ago

Help its a test and I’m pretty bad at these

saw5 [17]

B) just learn that you move the decimal right as numbers go higher and left if numbers decrease

So 1.75 km =1750000

Move decimal to the left to get a scientific notation

So 1.75 x 10^6

Brainliest?

8 0

2 years ago

Read 2 more answers

Consider the reaction: A <=> B. Under standard conditions at equiliubrium, the concentrations of the compounds are [A] = 1

Katen [24]

Answer:

$Keq'>1\\\Delta G'$

Explanation:

Hello,

In this case, for the given reaction, the equilibrium constant turns out:

$Keq=\frac{[B]}{[A]}=\frac{0.5M}{1.5M} =1/3$

Nonetheless, we are asked for the reverse equilibrium constant that is:

$Keq'=\frac{1}{Keq}=3$

Which is greater than one.

In such a way, the Gibbs free energy turns out:

$\Delta G'=-RTln(Keq')\\$

Now, since the reverse equilibrium constant is greater than zero its natural logarithm is positive, therefore with the initial minus, the Gibbs free energy is less than zero, that is, negative.

7 0

3 years ago

Read 2 more answers

A 6.10 M NaCl can be made by adding [x]g of NaCl to a container and making the volume of water up to the 1.00 L line

sergij07 [2.7K]

Answer:

356.484 g.

Explanation:

Molarity (M) is defined as the no. of moles of solute (NaCl) dissolved in a 1.0 L of the solvent.

<em>M = (no. of moles of NaCl)/(volume of the solution (L))</em>

M = 6.10 M, volume of the solution = 1.0 L.

∴ No. of moles of NaCl = (M)(volume of the solution (L)) = (6.10 M)(1.0 L) = 6.10 mol.

<em>∵ no. of moles of NaCl = (mass of NaCl)/(molar mass of NaCl)</em>

∴ Mass of NaCl = (no. of moles of NaCl)(molar mass of NaCl) = (6.10 mol)(58.44 g/mol) = 356.484 g.

8 0

2 years ago

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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

2 years ago