Isn’t urgent the first 3 will be marked BRAINIEST

Chemistry

2 answers:

valentinak56 [21]3 years ago

8 0

Aluminum Sulphate:

Al2(SO4)3

Magnesium Hydroxide:

Mg(OH)2

bearhunter [10]3 years ago

7 0

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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Pepsi [2]

Explanation:

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2 years ago

Suppose of copper(II) acetate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of acetate

uranmaximum [27]

Answer:

0.0714 M for the given variables

Explanation:

The question is missing some data, but one of the original questions regarding this problem provides the following data:

Mass of copper(II) acetate: $m_{(AcO)_2Cu} = 0.972 g$

Volume of the sodium chromate solution: $V_{Na_2CrO_4} = 150.0 mL$

Molarity of the sodium chromate solution: $c_{Na_2CrO_4} = 0.0400 M$

Now, when copper(II) acetate reacts with sodium chromate, an insoluble copper(II) chromate is formed:

$(CH_3COO)_2Cu (aq) + Na_2CrO_4 (aq)\rightarrow 2 CH_3COONa (aq) + CuCrO_4 (s)$

Find moles of each reactant. or copper(II) acetate, divide its mass by the molar mass:

$n_{(AcO)_2Cu} = \frac{0.972 g}{181.63 g/mol} = 0.0053515 mol$

Moles of the sodium chromate solution would be found by multiplying its volume by molarity:

$n_{Na_2CrO_4} = 0.0400 M\cdot 0.1500 L = 0.00600 mol$

Find the limiting reactant. Notice that stoichiometry of this reaction is 1 : 1, so we can compare moles directly. Moles of copper(II) acetate are lower than moles of sodium chromate, so copper(II) acetate is our limiting reactant.

Write the net ionic equation for this reaction:

$Cu^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow CuCrO_4 (s)$