Why do ionic compounds conduct when dissolved and covalent compounds do not?

A chemist must prepare 900.0mL of sodium hydroxide solution with a pH of 13.90 at 25°C. She will do this in three steps: Fill a

Ierofanga [76]

Answer:

28.58 g of NaOH

Explanation:

The question is incomplete. The missing part is:

<em>"Calculate the mass of sodium hydroxide that the chemist must weigh out in the second step"</em>

To do this, we need to know how much of the base we have to weight to prepare this solution.

First we know that is a sodium hydroxide aqueous solution so, this will dissociate in the ions:

NaOH -------> Na⁺ + OH⁻

As NaOH is a strong base, it will dissociate completely in solution, so, starting with the pH we need to calculate the concentration of OH⁻.

This can be done with the following expression:

14 = pH + pOH

and pOH = -log[OH⁻]

So all we have to do is solve for pOH and then, [OH⁻]. To get the pOH:

pOH = 14 - 13.9 = 0.10

[OH⁻] = 10⁽⁻⁰°¹⁰⁾

[OH⁻] = 0.794 M

Now that we have the concentration, let's calculate the moles that needs to be in the 900 mL:

n = M * V

n = 0.794 * 0.9

n = 0.7146 moles

Finally, to get the mass that need to be weighted, we need to molecular mass of NaOH which is 39.997 g/mol so the mass:

m = 39.997 * 0.7146

3 0

4 years ago

Consider the interaction of two hydrogen ls atomic orbitals of the same phase. Which statements below is an incorrest descriptio

Andreyy89

Answer:

B) The molecular orbital formed is lower in energy than a hydrogen 1s atomic orbital.

Explanation:

When two atoms of hydrogen come close to each other , there is formation of molecular orbital . Due to overlap of 1 s orbital of one and 1 s orbital of another atom , two molecular orbitals are formed . One of these molecular orbital has energy less than 1 s atomic orbital . It is called 1 s sigma bonding molecular orbital . The other molecular orbital has energy more than 1 s atomic orbital . It is called antibonding molecular orbital . Two electrons occupy bonding sigma molecular orbital .

So , the statement that "the molecular orbital formed is lower in energy than a hydrogen 1s atomic orbital " is wrong .

4 0

3 years ago

If you obtained 2.50 moles of CuSO4 from the reaction of CuO and H2SO4, but the theoretical yield was 3.19 moles, what was the p

IceJOKER [234]

Answer:

78.4%

Explanation:

Just as context, write the chemical equation and the mole ratios

1) <u>Balanced chemical equation</u>:

CuO (s) + H₂SO₄ (aq) → CuSO₄ (aq) + H₂O (l)

2) <u>Therotetical (stoichiometric) mole ratios</u>:

1 mol CuO : 1 mol H₂SO₄ : 1 mol CuSO₄ : 1 mol H₂O

You can calculate the percent yield from the amount of CuSO₄ obtained and the theoretical yield

3) <u>Percent yield</u>

Definition:

Percent yield = (actual yield / theoretical yield)×100

Theoretical yiedl (given): 3.19 moles CuSO₄

Actual yield (given): 2.50 moles CuSO₄

Calcualtion:

Substitute the values in the formula:

Percent yield = (2.50 moles CuSO₄ / 3.19 moles CuSO₄)×100 = 78.4%

6 0

4 years ago

Suppose a 500.mL flask is filled with 1.0mol of CO, 1.5mol of H2O and 0.70mol of CO2. The following reaction becomes possible: +

valentina_108 [34]

Answer:

[CO] = 0.62 M

Explanation:

Step 1: Data given

Volume of the flask = 500 mL

Number of moles CO = 1.0 moles

Number of moles H2O = 1.5 moles

Number of moles CO2 = 0.70 moles

The equilibrium constant K for this reaction is 3.80

Step 2: The balanced equation

CO(g) + H2O(g) ⇆ CO2(g) +H2(g)

Step 3: Calculate the initial concentrations

Concentration = moles / volume

[CO] = 1.0 moles / 0.500 L = 2.0 M

[H2O] = 1.5 moles / 0.500 L = 3.0 M

[CO2] = 0.70 moles / 0.500 L = 1.4 M

[H2] = 0M

Step 4: The concentration at the equilibrium

For 1 mol CO we have 1 mol H2O to produce 1 mol CO2 and 1 mol H2

[CO] = 2.0 -X M

[H2O] = 3.0 - X M

[CO2] =1.4 + X M

[H2] = X M

Step 5: Define Kc

Kc = [CO2][H2]/ [CO][H2O]

3.80 = (1.4 + X) * X / ((2.0 - X)*3.0 -X))

X = 1.38

[CO] = 2.0 -1.38 = 0.62 M

[H2O] = 3.0 - 1.38 = 1.62 M

[CO2] =1.4 + 1.38 M = 2.78

[H2] = 1.38 M

Kc = (2.78*1.38) / (0.62*1.62)

Kc = 3.8

[CO] = 0.62 M

8 0

3 years ago

The metal zirconium becomes superconducting at temperatures below 3.4000K.

dybincka [34]

Answer:

-269.75 degrees Celsius

Explanation:

7 0

3 years ago