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3 years ago

14

An atom has the following electron configuration . 1s^ 2 2s^ 2 2p^ 6 3s^ 2 3p^ 6 4s^ 2 How many valence electrons does this atom

have ?
2
3
5
11

1 answer:

stepan [7]3 years ago

5 0

It’s 2 because it’s the number of row it is in; the element is calcium in row 2

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frozen [14]

Characteristic properties are used because the sample size and the shape of the substance does not matter.

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3 years ago

What is Industrial chemistry?

mel-nik [20]

Answer:

study of chemical used in industries

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3 years ago

S<br> SIO₂ +3C Sic +20o<br> Sic ? mass in grams<br> С c-

seraphim [82]

Answer:

26.74g

Explanation:

The equation of the reaction is;

SIO₂ + 3C --> SiC +2CO

From the balanced equation, the relationship between SiC and C is;

3 mol of C produces 1 mol of SiC

Converting mol to mass using; Mass = moles * Molar mass

Mass of SiC = 1 mol * 40.11 g/mol = 40.11g

This means;

3 mol of C produces 40.11g of SiC

2 mol of C produces xg of SiC

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2 = x

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6 0

3 years ago

lubasha [3.4K]

Answer:

2 circles one proton and one nucleon.draw quarks within each. strong nuclear force within protons between quarks and residual strong force between proton and nucleon (up,up,down in proton)

Explanation:

6 0

3 years ago

The Handbook of Chemistry and Physics gives solubilities of the following compounds in grams per 100 mL water. Because these com

Elanso [62]

Answer:

(a) $Ksp=4.50x10^{-7}$

(b) $Ksp=1.55x10^{-6}$

(c) $Ksp=2.27x10^{-12}$

(d) $Ksp=1.05x10^{-22}$

Explanation:

Hello,

In this case, given the solubility of each salt, we can compute their molar solubilities by using the molar masses. Afterwards, by using the mole ratio between ions, we can compute the concentration of each dissolved and therefore the solubility product:

(a) $BaSeO_4(s)\rightleftharpoons Ba^{2+}(aq)+SeO_4^{2-}(aq)$

$Molar\ solubility=\frac{0.0188g}{100mL} *\frac{1mol}{280.3g}*\frac{1000mL}{1L}=6.7x10^{-4}\frac{mol}{L}$

In such a way, as barium and selenate ions are in 1:1 molar ratio, they have the same concentration, for which the solubility product turns out:

$Ksp=[Ba^{2+}][SeO_4^{2-}]=(6.7x10^{-4}\frac{mol}{L} )^2\\\\Ksp=4.50x10^{-7}$

(B) $Ba(BrO_3)_2(s)\rightleftharpoons Ba^{2+}(aq)+2BrO_3^{-}(aq)$

$Molar\ solubility=\frac{0.30g}{100mL} *\frac{1mol}{411.15g}*\frac{1000mL}{1L}=7.30x10^{-3}\frac{mol}{L}$

In such a way, as barium and bromate ions are in 1:2 molar ratio, bromate ions have twice the concentration of barium ions, for which the solubility product turns out:

$Ksp=[Ba^{2+}][BrO_3^-]^2=(7.30x10^{-3}\frac{mol}{L})(3.65x10^{-3}\frac{mol}{L})^2\\\\Ksp=1.55x10^{-6}$

(C) $NH_4MgAsO_4(s)\rightleftharpoons NH_4^+(aq)+Mg^{2+}(aq)+AsO_4^{3-}(aq)$

$Molar\ solubility=\frac{0.038g}{100mL} *\frac{1mol}{289.35g}*\frac{1000mL}{1L}=1.31x10^{-4}\frac{mol}{L}$

In such a way, as ammonium, magnesium and arsenate ions are in 1:1:1 molar ratio, they have the same concentrations, for which the solubility product turns out:

$Ksp=[NH_4^+][Mg^{2+}][AsO_4^{3-}]^2=(1.31x10^{-4}\frac{mol}{L})^3\\\\Ksp=2.27x10^{-12}$

(D) $La_2(MoOs)_3(s)\rightleftharpoons 2La^{3+}(aq)+3MoOs^{2-}(aq)$

$Molar\ solubility=\frac{0.00179g}{100mL} *\frac{1mol}{1136.38g}*\frac{1000mL}{1L}=1.58x10^{-5}\frac{mol}{L}$

In such a way, as the involved ions are in 2:3 molar ratio, La ion is twice the molar solubility and MoOs ion is three times it, for which the solubility product turns out:

$Ksp=[La^{3+}]^2[MoOs^{-2}]^3=(2*1.58x10^{-5}\frac{mol}{L})^2(3*1.58x10^{-5}\frac{mol}{L})^3\\\\Ksp=1.05x10^{-22}$

Best regards.

7 0

4 years ago