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3 years ago

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In the following reaction: Mg + 2HCl → MgCl2 + H2 How many liters of H2 would be produced if you started with 24.3 g of Mg?

1 answer:

Aloiza [94]3 years ago

4 0

Answer:

22.4 L H2

Explanation:

There is a better explanation brainly.com/question/9562878

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G determine the concentration of an hbr solution if a 45.00 ml aliquot of the solution yields 0.6485 g agbr when added to a solu

Sunny_sXe [5.5K]

The molecular weight of silver bromide (AgBr) is 187.77 g/mole. The presence of the ions in solution can be shown as- AgBr (insoluble) ⇄ $Ag^{+}$ + $Br^{-1}$ .

45.00 mL of the aliquot contains 0.6485 g of AgBr. Thus 1000 mL of the aliquot contains $\frac{0.6485}{45}$ ×1000 = 14.411 gm-mole. Thus the solubility product $K_{sp}$ of AgBr = [ $Ag^{+}$ ]× $Br^{-}$ .

Or, 5.0× $10^{-13}$ = $S^{2}$ (the given value of solubility product of AgBr is 5.0× $10^{-13}$ and the charge of the both ions are same).

Thus S = (5.00× $10^{-13}$ ) $^{1/2}$ = 7.071× $10^{-7}$ g/mL.

Thus the concentration of Br $^{-1}$ or HBr is 7.071× $10^{-7}$ g/mL.

4 0

3 years ago

La masa de una olla es de 300g y contiene 90% de aluminio. Hallar el número de moles de aluminio de la olla. P.A.(Al= 27)

Neko [114]

Explanation:

The mass of a pot is 300g and contains 90% aluminum. Find the number of moles of aluminum in the pot. P.A. (Al = 27)

The mass of aluminum present in the pot is:

$300 g * 90/100\\=270 g$

Hence, in the given pot 270g Al is present.

$Number of moles of Al=\frac{given mass ofAl}{its molar mass}$

The gram atomic mass of Al -27 g/mol

Given the mass of Al is 270 g

Substitute these values in the above formula:

$Number of moles of Al=\frac{given mass ofAl}{its molar mass}\\=\frac{270 g}{27 g} \\=10.0 mol$

Answer is 10.0 mol of Al is present.

6 0

2 years ago

Isn’t urgent the first 3 will be marked BRAINIEST

bearhunter [10]

Answer:

Al2(SO4)3 and Mg(OH)2

Explanation:

1. Al has a charge of 3-, and SO4 of 2-

when you cross multiply the charges you get

Al2 and (SO4)3

*the reason theres a bracket around the sulfate ion is that the charge 3 is not for oxygen only, but the entire sulphate ion*

Hence, Al2(SO4)3

2. Mg has a charge of 2- and OH of 1-

again cross multiply

Mg (you dont need to add the 1) and (OH)2

again, the bracket around OH means the charge appiles to Oxygen AND hydrogen

hence, Mg(OH)2

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