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3 years ago

How many protons, electrons, and neutrons does the following isotope contain? ^{18} \text{F}^{-} 18 F −

1 answer:

8 0

Isotope ¹⁸F⁻ contains:
1) p⁺ = 9; number of protons.
Fluorine has a<span>tomic number Z = 9 (total number of protons).
2) e</span>⁻<span> = 10; </span>number of electrons.<span>
In element number of electrons and protons are the same, because element has neutral charge, but because in this example, fluorine is anion with negative charge, it has one electron more.
3) n</span>° = 9; number of neutrons.
<span>Mass number A = 18 is total number of protons and neutrons in a nucleus, so number of neutrons is A-Z = 18-9=9.</span>

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Solid aluminum and gaseous oxygen read in a combination reaction to produce aluminum oxide. 4Al(s) + 3O_2(g) rightarrow 2Al_2O_3

Kryger [21]

Answer:

4.7 g. Option 5 is the right one.

Explanation:

4Al(s) + 3O₂ (g) ⇄ 2Al₂O₃ (s)

We convert the mass of reactants to moles, in order to determine the limiting.

2.5 g Al / 26.98 g/mol → 0.092 moles of Al

2.5 g O₂ / 32g/mol → 0.078 moles of O₂

Ratio is 4:3. 4 moles of Al react with 3 moles of O₂

Then, 0.092 moles of O₂ would react with (0.092 . 3)/ 4 = 0.069 moles O₂

We have 0.078 moles of O₂ and we need 0.069 moles, the oxygen is the limiting in excess. Therefore the Al is the limiting reactant.

Ratio is 4:2. 4 moles of Al, can produce 2 moles of Al₂O₃

Then, 0.092 moles of Al would produce (0.092 .2) / 4 = 0.046 moles

If we convert the moles to mass, we find the anwer:

0.046 mol . 101.96 g/mol = 4.69 g

5 0

3 years ago

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REY [17]

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3 years ago

6. A solution of CuSO4 is electrolyzed for 10 minutes with a current of 1.5 amperes. What is the mass of copper deposited at the

vladimir2022 [97]

The mass of Copper deposited at the cathode : 0.296 g

<h3>Further explanation</h3>

Given

time = t = 10 min=600 s

current = i = 1.5 A

F = 96500 C

charge Cu=+2

Required

The mass of Copper

Solution

Faraday's Law

$\tt W=\dfrac{e.i.t}{96500}$

e = Ar/valence(valence Cu=2, Ar=63.5 g/mol)

Input the value :

$\tt W=\dfrac{63.5/2\times 1.5\times 600}{96500}\\\\W=0.296~g$

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Kay [80]

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stealth61 [152]

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