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Aleksandr [31]
3 years ago
13

A neutral aluminum atom is ionized and loses 3 electrons. What is the

Chemistry
1 answer:
Nata [24]3 years ago
6 0

Explanation:

The neutral aluminium atom loses all 3 of its valence electrons to obtain a stable octet structure.

The aluminium ion has a positive charge of 3.

=> Al3+, Charge = +3.

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A sample of aspirin weighing 5.945 g became contaminated with 2.134 g of sodium sulfate. The resulting
ira [324]

Answer:

The correct answer to the following question will be "62.9 %".

Explanation:

The given values are:

The aspirin's initial amount = 5.945 g.

and is polluted containing 2,134 g of sodium sulfate.

After extraction we provided 3,739 g of pure aspirin.

Now,

Percentage \ of \ recovery \ aspirin =\frac{Aspirin \ amount \ isolated \ after \ extraction}{initial \ amount \ of \ aspirin}\times 100

On putting the values in the above formula, we get

⇒                                                    =\frac{3.739}{5.945}\times 100

⇒                                                    =62.9 \ percent

Note: percent = %

7 0
3 years ago
Help............???????
balandron [24]

B. .175


Hope it helps!

4 0
3 years ago
Sulfuric acid is produced in larger amounts by weight than any other chemical. It is used in manufacturing fertilizers, oil refi
Fed [463]

Answer:

A. -166.6 kJ/mol

B. -127.7 kJ/mol

C. -133.9 kJ/mol

Explanation:

Let's consider the oxidation of sulfur dioxide.

2 SO₂(g) + O₂(g) → 2 SO₃(g)     ΔG° = -141.8 kJ

The Gibbs free energy (ΔG) can be calculated using the following expression:

ΔG = ΔG° + R.T.lnQ

where,

ΔG° is the standard Gibbs free energy

R is the ideal gas constant

T is the absolute temperature (25 + 273.15 = 298.15 K)

Q is the reaction quotient

The molar concentration of each gas ([]) can be calculated from its pressure (P) using the following expression:

[]=\frac{P}{R.T}

<em>Calculate ΔG at 25°C given the following sets of partial pressures.</em>

<em>Part A  130atm SO₂, 130atm O₂, 2.0atm SO₃. Express your answer using four significant figures.</em>

[SO_{2}]=[O_{2}]=\frac{130atm}{(0.08206atm.L/mol.K).298K} =5.32M

[SO_{3}]=\frac{2.0atm}{(0.08206atm.L/mol.K).298K} =0.0818M

Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0818^{2} }{5.32^{3} } =4.44 \times 10^{-5}

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln (4.44 × 10⁻⁵) = -166.6 kJ/mol

<em>Part B  5.0atm SO₂, 3.0atm O₂, 30atm SO₃  Express your answer using four significant figures.</em>

<em />

[SO_{2}]=\frac{5.0atm}{(0.08206atm.L/mol.K).298K}=0.204M

[O_{2}]=\frac{3.0atm}{(0.08206atm.L/mol.K).298K}=0.123M

[SO_{3}]=\frac{30atm}{(0.08206atm.L/mol.K).298K}=1.23M

Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{1.23^{2} }{0.204^{2}.0.123 } =296

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 296 = -127.7 kJ/mol

<em>Part C Each reactant and product at a partial pressure of 1.0 atm.  Express your answer using four significant figures.</em>

<em />

[SO_{2}]=[O_{2}]=[SO_{3}]=\frac{1.0atm}{(0.08206atm.L/mol.K).298K}=0.0409M

Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0409^{2} }{0.0409^{3}} =24.4

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 24.4 = -133.9 kJ/mol

7 0
3 years ago
How many neutrons are probably in the nucleus of an element of atomic weight 197? 
Ratling [72]
Atomic weight = 197
symbol = Au
electrons = 79

neutrons = 197 - 79 = 118

<u>answer: E</u>
4 0
3 years ago
A solution in which more solute can be dissolved is called a(n) ???? solution
Svet_ta [14]

Answer:it's called unsaturated solution, because that's it's definition

8 0
3 years ago
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