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2 years ago

Question 4

Chemistry

1 answer:

kirill [66]2 years ago

4 0

Answer:

9.33

Explanation:

balanced chemical equation:

2NH₃ + 3CuO → 3Cu + N₂ + 3H₂O

14 moles of NH3

2NH₃ → 3H₂O

14 * 2/3 = 9.33

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20 grams of H₂S react with 20 grams of O₂. What is the limiting reagent? *

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3 years ago

If you mixed three Dixie cups’ contents together containing 0.05L of 1.0M lemonade, 0.05L of 2.5M lemonade, and 0.05L of 0.5M le

viva [34]

Molarity of the resulting solution will be 1.33 M.

Explanation:

First we have to find the number of moles for each of the solution using the formula, moles = molarity × volume

For cup 1 = 1 M ×0.05 L = 0.05 moles

For cup 2 = 2.5 M × 0.05 L= 0.125 moles

For cup 3 = 0.5 M × 0.05 L = 0.025 moles

Total moles = 0.05 + 0.125 + 0.025 = 0.2 moles

We have to find the total volume as, 0.05 + 0.05 + 0.05 = 0.15 L

Now we have to find the molarity as, moles / volume = 0.2 moles/ 0.15 L = 1.33 M

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3 years ago

Can someone check my answers? Identify the structure as having hydrogen bonding, dipole-dipole moments or London dispersion forc

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4 years ago

Which of the following chemical compounds contains two hydrogen atoms in every molecule?

mina [271]

Answer:

A. Water

Explanation:

H2O means that there are two hydrogen and one oxygen in every water molecule.

7 0

4 years ago

A 35.0 mL sample of 0.150 M acetic acid (CH3COOH) is titrated with 0.150 M NaOH solution. Calculate the pH after the following v

marissa [1.9K]

Answer:

a) 2.79

b) 4.75

c) 6.59

d) 8.81

e) 11.03

f) 12.42

Explanation:

Step 1: Data given

Volume of acetic acid = 35.0 mL

Molarity of acetic acid = 0.150 M

Molarity of NaOH = 0.150 M

Step 2: pH after adding 0 mL of NaOH

HA + H2O ⇆ A- + H3O+

Initial concentration

[HA] = 0.150 M

[A-] = 0M

[H3O+] = 0M

Concentration at the equilibrium

[HA] = 0.150 - x M

[A-] = xM

[H3O+] = xM

Ka = [H3O+][A-] / [HA] = x²/(0.150 -x) = 1.76*10^-5

0.150 >>> x

We can write it like this

x²/(0.150) = 1.76*10^-5

x² =0.00000264

x = 0.00162

[H3O+] = 0.00162 M

pH = -log [H3O+] = -log (0.00162) = 2.79

Step 3: pH after adding 17.5 mL of NaOH

This is the point of half-neutralization.

pH = pKa = -log (1.76*10^-5- 4.75

Step 4: pH after adding 34.5 mL of NaOH

HA + OH- ⇆ A- + H2O

Moles acetic acid = molarity * volume

Moles acetic acid = 0.150 M * 0.035 L

Moles acetic acid = 0.00525 moles

Moles NaOH = 0.150 M *0.0345 L

Moles NaOH = 0.005175 moles

Initial moles

[HA] = 0.00525 moles

[OH-] = 0.005175 moles

[A-] = 0 moles

[H2O] = 0 moles

Moles at the equilibrium

[HA] = 0.00525 - 0.005175 = 0.000075 moles

[OH-] = 0.005175 - 0.005175 = 0 moles

[A-] = 0.005175 moles

pH = pKa + log([A-]/[HA]

pH = 4.75 +log(0.005175/0.000075)

pH = 4.75 + 1.84 = 6.59

Step 4: pH after adding 35.0 mL of NaOH

HA + OH- ⇆ A- + H2O

Moles acetic acid = molarity * volume

Moles acetic acid = 0.150 M * 0.035 L

Moles acetic acid = 0.00525 moles

Moles NaOH = 0.150 M *0.035 L

Moles NaOH = 0.00525 moles

Initial moles

[HA] = 0.00525 moles

[OH-] = 0.00525 moles

[A-] = 0 moles

[H2O] = 0 moles

Moles at the equilibrium

[HA] = 0.00525 - 0.00525 = 0 moles

[OH-] = 0.00525 - 0.00525 = 0 moles

[A-] = 0.00525 moles

We have a solution of 0.00525 mol of A - in 70.0 mL

[A-] = 0.00525 moles / 0.070 L = 0.075 M

A- + H20 ⇆ HA + OH-

Initial moles A- = 0.0750 moles

initial moles HA = 0 moles

Initial moles OH- = 0 moles

Moles A- at the equilibrium = 0.0750 - x

Moles HA at the equilibrium = x

Moles OH- at the equilibrium = x

Kb = Kw/Ka = 10^-14 / 1.76*10^-5

Kb = 5.68 *10^-10

Kb = [HA][OH-] / [A-] = x² / 0.0750 - x = 5.68*10^-10

Since 0.0750 >>> x we can write

5.68*10^-10 = x²/0.0750

x = 6.53 *10^-6

[OH-] = x = 6.53 * 10^-6 M

pOH = - log (6.53*10^-6) = 5.19

pH = 14 - pOH = 14 - 5.19 = 8.81

Step 5: pH after adding 35.5 mL of NaOH

We add an excess moles of NaOH

We have an excess of 0.5 mL NaOH

Moles excess = 0.0005 L *0.150 M = 0.000075 moles

[OH-] = 0.000075 moles / (0.035+0.0355L)

[OH-] = 1.06 * 10^-3 M

pOH = -log( 1.06 *10^-3)

pOH = 2.97

pH = 14 -2.97 = 11.03

Step 6: pH after adding 50.0 mL of NaOH

We have an excess of 15.0 mL NaOH

Moles excess = 0.015 L *0.150 M = 0.00225 moles

[OH-] = 0.00225 moles / (0.035+0.050L)

[OH-] = 0.0265 M

pOH = -log( 0.0265M)

pOH = 1.58

pH = 14 -1.58 = 12.42

4 0

3 years ago