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3 years ago

8

A member of the boron family has three valence electrons, while a member of the nitrogen family has (1 point)none. five. four. t

hree.

1 answer:

Daniel [21]3 years ago

4 0

Answer: lesson 11 practice

1. Not part-time contain subatomic particles

2. Nucleus in atom-Rutherford

3. Bee hovering around give

4. Mass

5. Atomic mass unit

6. 5

7. Li2O

8. Strong attractions between polar water molecules

Explanation:

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Na2SO4<br> How many oxygen stems are?

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Answer:

Na2SO4 means: two moles sodium (45.98 g), one mole sulfur (32.06 g), and four moles oxygen (64.00 g) combine to form one mole of sodium sulfate (142.04 g).

Explanation:

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3 years ago

Which of the following is a precipitation reaction?

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Answer:

B is a precipitation reaction.

Explanation:

This is because a precipitation reaction is when a solid is made from the combination of cations and anions in a solution to create a solid.

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Write a hypothesis:<br> What is the mole ratio of the cation and the anion in a precipitate?

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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

4 years ago