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3 years ago

Why does Al have a larger atomic radius than B

Chemistry

1 answer:

Valentin [98]3 years ago

8 0

Explanation:

While atomic radii DECREASE across a Period, a row of the Period Table, from left to right as we face the Table, atomic radii INCREASE down a Group, a column of the Periodic Table, due to the shielding of the valence electrons by the closed valence shell(s) that intervenes between the nucleus and the valence, outermost electrons.

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A sample of coal has the following analysis (wt %). Moisture 1.1%, Fixed Carbon 74%, Volatile Matter 17.9%, Carbon 63.7%, Hydrog

pogonyaev

Answer:

Coal is a traditionally used source of energy, there are main four types of ranks for coal. Here the rank of a coal means to a natural process called Coalification, which takes place during a plant is buried and changes to a harder, and denser material and become even more rich in carbon contents.

Anthracite is know to have the highest ranked coal, it contains highest percent of fixed carbon and lowest percent of volatile material.

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3 years ago

Ad libitum [116K]

Explanation:

No because the person on the bike is going faster then the person on foot.

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3 years ago

Predict the formula for a compound made from X^2+ and Y^3-

slamgirl [31]

Compound is X3Y2.

Hope this helps you!

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3 years ago

For the reaction 2NH3(g) + 2O2(g)N2O(g) + 3H2O(l) H° = -683.1 kJ and S° = -365.6 J/K The standard free energy change for the rea

BlackZzzverrR [31]

Answer:

$\Delta G^{0}$ = -457.9 kJ and reaction is product favored.

Explanation:

The given reaction is associated with 2 moles of $NH_{3}$

Standard free energy change of the reaction ( $\Delta G^{0}$ ) is given as:

$\Delta G^{0}=\Delta H^{0}-T\Delta S^{0}$ , where T represents temperature in kelvin scale

So, $\Delta G^{0}=(-683.1\times 10^{3})J-(273K\times -365.6J/K)=-583291.2J$

So, for the reaction of 1.57 moles of $NH_{3}$ , $\Delta G^{0}=(\frac{1.57}{2})\times -583291.2J=-457883.592J=-457.9kJ$

As, $\Delta G^{0}$ is negative therefore reaction is product favored under standard condition.

6 0

3 years ago

Given 4.80g of ammonium carbonate, find:

V125BC [204]

Answer:

1) 0.05 mol.

2) 0.1 mol.

3) 0.05 mol.

4) 0.4 mol.

5) 2.4 x 10²³ molecules.

Explanation:

1) Number of moles of the compound:

no. of moles of ammonium carbonate = mass/molar mass = (4.80 g)/(96.09 g/mol) = 0.05 mol.

2) Number of moles of ammonium ions :

Ammonium carbonate is dissociated according to the balanced equation:

(NH₄)₂CO₃ → 2NH₄⁺ + CO₃²⁻.

It is clear that every 1.0 mole of (NH₄)₂CO₃ is dissociated to produce 2.0 moles of NH₄⁺ ions and 1.0 mole of CO₃²⁻ ions.

∴ The no. of moles of NH₄⁺ ions in 0.05 mol of (NH₄)₂CO₃ = (2.0)(0.05 mol) = 0.1 mol.

3) Number of moles of carbonate ions :

Ammonium carbonate is dissociated according to the balanced equation:

(NH₄)₂CO₃ → 2NH₄⁺ + CO₃²⁻.

It is clear that every 1.0 mole of (NH₄)₂CO₃ is dissociated to produce 2.0 moles of NH₄⁺ ions and 1.0 mole of CO₃²⁻ ions.

∴ The no. of moles of CO₃²⁻ ions in 0.05 mol of (NH₄)₂CO₃ = (1.0)(0.05 mol) = 0.05 mol.

4) Number of moles of hydrogen atoms:

Every 1.0 mol of (NH₄)₂CO₃ contains:

2.0 moles of N atoms, 8.0 moles of H atoms, 1.0 mole of C atoms, and 3.0 moles of O atoms.

∴ The no. of moles of H atoms in 0.05 mol of (NH₄)₂CO₃ = (8.0)(0.05 mol) = 0.4 mol.

5) Number of hydrogen atoms:

It is known that every mole of a molecule or element contains Avogadro's number (6.022 x 10²³) of molecules or atoms.

Using cross multiplication:

1.0 mole of H atoms contains → 6.022 x 10²³ atoms.

0.4 mole of H atoms contains → ??? atoms.

∴ The no. of atoms in 0.4 mol of H atoms = (6.022 x 10²³ molecules)(0.4 mole)/(1.0 mole) = 2.4 x 10²³ molecules.

8 0

3 years ago