Type the correct answer in each box.

75.0 grams of MgCl, is dissolved in 500.0 g of water, density 1.00 g/ml. What is the MOLALITY of this solution?

ollegr [7]

Answer:

C

Explanation:

Check it. So u can know it

8 0

2 years ago

What is the quantum number set of the ground-state electron that is found in lithium (Li) but not in helium (He)?

statuscvo [17]

The electron configuration of lithium atom is: $Li:[He]2s^1$

The number "2" is the value of the principal quantum number "n". Letter "s" is associated with the value of secondary quantum number "l" and it is equal to zero. The value of "m" (or magnetic quantum number) is zero too. The quantum number set for the highest energy electron will be (2, 0, 0, 1/2).

3 0

3 years ago

What is the molarity of a NaOH solution if 28.2 mL of a 0.355 M H2SO4 solution is required to neutralize a 25.0-mL sample of the

Verdich [7]

Answer:

[NaOH} = 0.4 M

Explanation:

In a reaction of neutralization, we determine the equivalence point of the titration. In this case, we have a strong base and a strong acid.

(H₂SO₄, is considered strong, but the first deprotonation is weak)

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

As we have 2 protons in the acid, we need 2 OH⁻ from the base to form 2 molecules of water.

In the equivalence point we know mmoles of base = mmoles of acid

Let's finish the excersise with the formula

25 mL . M NaOH = 28.2 mL . 0.355M

M NaOH = (28.2 mL . 0.355M) / 25 mL → 0.400

8 0

3 years ago

A) Calculate the pH of a 2.0x10-4 M solution of aniline hydrochoride, C6H5NH3Cl.

igor_vitrenko [27]

Given the concentration of aniline hydrochloride is $2.0 *10^{-4} M$

Aniline hydrochloride is the conjugate acid of aniline a weak base.

pH can be calculated from $pK_{a}$ anilinium ion the conjugate acid of aniline.

8 0

3 years ago

Read 2 more answers

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