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AVprozaik [17]

1000 g of feathers is the least dense

6 0

3 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:

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

Notice that acetate is the ion spectator. This means it doesn't react, its moles throughout reaction stay the same. We started with:

$n_{(AcO)_2Cu} = 0.0053515 mol$

According to stoichiometry, 1 unit of copper(II) acetate has 2 units of acetate, so moles of acetate are equal to:

$n_{AcO^-} = 2\cdot 0.0053515 mol = 0.010703 mol$

The total volume of this solution doesn't change, so dividing moles of acetate by this volume will yield the molarity of acetate:

$c_{AcO^-} = \frac{0.010703 mol}{0.1500 L} = 0.0714 M$

8 0

3 years ago

Which pair of samples contains the same number of oxygen atoms in each compound?0.20 mol Ba(OH)2 and 0.20 mol H2SO40.20 mol Br2O

Whitepunk [10]

Answer:

0.20 mol Br2O and 0.20 mol HBrO have the same number of oxygen atoms

Explanation:

0.20 mol Ba(OH)2 and 0.20 mol H2SO4

In Ba(OH)2 there are 2 moles of O atoms in every mol of Ba(OH)2.

Number of O atoms in 0.20 moles Ba(OH)2 = 2*0.20 = 0.40 moles O atom

In H2SO4 there are 4 moles of O atoms for every mol of H2SO4.

Number of O atoms in 0.20 moles H2SO4 = 4*0.20 = 0.80 moles O atom

⇒ 0.20 mol Ba(OH)2 and 0.20 mol H2SO4 do not have the same number of oxygen atoms.

0.20 mol Br2O and 0.20 mol HBrO

In Br2O there is 1 mol of O atoms in every mol Br2O

Number of O atoms in 0.20 moles Br2O = 0.20*1 = 0.20 moles O atom

In HBrO there is 1 mol of O atom in every mol HBrO

Number of O atoms in 0.20 moles HBrO = 0.20 *1 = 0.20 moles O atom

⇒ in 0.20 moles Br2O and 0.20 moles HBrO we have the same number of oxygen atoms

0.20 moles of Oygen contains: = 0.20 * 6.022*10^23 = 1.2 *10^23 O atoms

0.10 mol Fe2O3 and 0.50 mol BaO

In Fe2O3 there are 3 moles of O atoms in every mol of Fe2O3

Number of O atoms in 0.10 moles Fe2O3 = 0.10 * 3 = 0.30 moles O atom

In BaO there is 1 mol of O atoms in every mol BaO

Number of O atoms in 0.50 mol BaO = 1*0.50 = 0.50 moles O atom

⇒ 0.10 mol Fe2O3 and 0.50 mol BaO do not have the same number of oxygen atoms.

0.10 mol Na2O and 0.10 mol Na2SO4

In Na2O there is 1 mol of O atoms in every mol of Na2O

Number of O atoms in 0.10 mol Na2O = 1*0.10 = 0.10 moles O atom

In Na2sO4 there are 4 moles of O atoms in every mol of Na2SO4

Number of O atoms in 0.10 mol Na2SO4 = 4*0.10 = 0.40 moles O atom

⇒ 0.10 mol Na2O and 0.10 mol Na2SO4 do not have the same number of oxygen atoms.

4 0

3 years ago

Which chemical class does phenobarbital belong to?

drek231 [11]

Phenobarbital is derivative of Barbituric Acid and Barbituric Acid is derivative of Urea. (structures shown in Fig below)

Urea has H₂N- group attached to Carbonyl Group (C=O), and such class of comounds conataining H₂N-C=O bond are called as Amides.

Result:
So, Phenobarbital belongs to Amides.

3 0

3 years ago

When the chromosomes line up in mitosis this is know as which phase

Harrizon [31]

Metaphase

Hope this helps

5 0

3 years ago

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