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Viefleur [7K]
3 years ago
10

Which element or ion listed below has the electron configuration 1s22s22p6?

Chemistry
1 answer:
Anna11 [10]3 years ago
7 0
Count up the number of electrons in each orbital. theres ten electrons which means it has 10 protons. the element would be neon
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Which element is a halogen?<br><br> argon <br> bromine <br> calcium <br> lithium
almond37 [142]
It is (CI) bromine
because, <span>Halogen element, any of the six nonmetallic elements that constitute Group 17 (Group VIIa) of the periodic table. The halogen elements are </span>fluorine (F)<span>, </span>chlorine (Cl), bromine (Br<span>), iodine (I), astatine (At), and tennessine (Ts).</span>
6 0
3 years ago
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A mixture contains only nacl and al2(so4)3. a 1.45 g sample of the mixture is dissolved in water, and an excess of naoh is added
Aleks [24]

When mixture of NaCl and Al₂(SO₄)₃ is allowed to react with excess NaOH, only Al₂(SO₄)₃ reacts with it and NaCl does not react with NaOH due to presence of common ion (Na⁺). On reaction gelatinous precipitate of aluminium hydroxide [Al(OH)₃] is produced.  The balanced chemical reaction is represented as-

Al₂(SO₄)₃ + 6NaOH → 2Al(OH)₃ + 3Na₂SO₄

On this reaction, 0.495 g = 0.495/78 moles =6.346 X 10⁻³ moles of Al(OH)₃.

As per balanced reaction, two moles of Al(OH)₃ is produced from one mole Al₂(SO₄)₃. So, 6.346 X 10⁻³ moles of Al(OH)₃ is produced from (6.346 X 10⁻³)/2 moles=3.173 X 10⁻³ moles of Al₂(SO₄)₃= 3.173 X 10⁻³ X 342 g of Al₂(SO₄)₃=1.085 g of Al₂(SO₄)₃.

So, mass percentage of Al₂(SO₄)₃ is= (amount of Al₂(SO₄)₃/total amount of mixture)X100 = \frac{1.085 X 100}{1.45} =74.8 %.

7 0
3 years ago
When a solution of lead(II) nitrate is mixed with a solution of sodium chromate, a yellow precipitate forms. (a) Enter the balan
Damm [24]

Answer:

a) Pb(NO₃)₂(aq) + Na₂Cr₂O₄(aq) ⇄ 2NaNO₃(aq) + Pb(Cr₂O₄)(s)

b) 67.6%

Explanation:

a) Nitrate is the ion NO₃⁻, and lead(II) forms the ion Pb⁺², so the compound lead(III) nitrate is Pb(NO₃)₂. (First, the cation, then the anion, with charges replaced without the signal).

Chromate is the ion Cr₂O₄⁻² and sodium forms the ion Na⁺, so the sodium chromate is Na₂Cr₂O₄. Both of them are in solutions, so they will be in an aqueous state.

In the reaction, the anions and cations will replace and will form: NaNO₃ and Pb(Cr₂O₄). The nitrates formed by metals from group 1, such as sodium, are soluble, so it will not forme a precipitated. So, the precipitated is PbCr₂O₄, and the balanced reaction is:

Pb(NO₃)₂(aq) + Na₂Cr₂O₄(aq) ⇄ 2NaNO₃(aq) + Pb(Cr₂O₄)(s)

b) The molar masses are: Pb(NO₃)₃ = 331,2 g/mol; Na₂Cr₂O₄ = 162 g/mol; Pb(Cr₂O₄) = 323,2 g/mol.

First, let's find what is the limiting reactant, doing the stoichiometry calculus for the reactants. Let's suppose that Na₂Cr₂O₄ is the limiting so:

1 mol of Pb(NO₃)₂ ------------------------------ 1 mol of Na₂Cr₂O₄

Transforming to mass (mass = moles * molar mass):

331,2 g of Pb(NO₃)₂ ------------------------- 162 g/mol of Na₂Cr₂O₄

x ------------------------- 12.38

By a simple direct three rule:

162x = 4100.256

x = 25.3 g of Pb(NO₃)₂

This is higher than what is put in the reaction, so Pb(NO₃)₂ is the limiting reactant, and Na₂Cr₂O₄ is in excess. So, let's do the stoichiometric calculus for the limiting reactant and the solid formed:

1 mol of Pb(NO₃)₂ ----------------------- 1 mol of Pb(Cr₂O₄)

Transforming to mass:

331.2 g of Pb(NO₃)₂ ------------------- 323.2 g of Pb(Cr₂O₄)

11.39 g ------------------- y

By a simple direct three rule:

331.2 y = 3681.248

y = 11.115 g

The yield is the mass formed divided by the stoichiometric result multiplied by 100%:

yield = (7.52/11.115)*100% = 67.6%

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Umm. . . Could you list some options? :)
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