1. List five acids and five bases we come across in our day-to-day activities under the correct heading.

Name the following bases.

zheka24 [161]

Answer:

1

Explanation:

Ferric Hydroxide

3. 1

Fe. (OH)

Fe(OH)3

6 0

2 years ago

When the isotonic saline solution described in part A is connected to an unknown solution via a semipermeable membrane, the unkn

ozzi

B. The unknown solution had the lower concentration.

Explanation:

Osmosis is a phenomenon in which the molecules of the solvent has a tendency to move through a membrane which is semipermeable from lower concentrated side to the higher concentration side, so that the concentrations on both sides of the membrane must be equal.

So the unknown solution may have lesser concentration than the isotonic solution so that molecules of that solution move from less concentrated side to the more concentrated side, so its level drops.

3 0

3 years ago

Why is propionic acid soluble in water

omeli [17]

HSDB and the SRC Physical Properties database database reports that propionic acid has a water solubility of 1000 g/l at 25 °C. Ullmann and Knovel Solvents - A Properties Database reports that propionic acid is miscible in any ratio with water.

4 0

3 years ago

Ammonia NH3 may react with oxygen to form nitrogen gas and water.4NH3 (aq) + 3O2 (g) \rightarrow 2 N2 (g) + 6H2O (l)If 2.15g of

bagirrra123 [75]

Answer:

NH3 is the limiting reactant

The % yield is 36.1 %

Explanation:

Step 1: Data given

Mass of NH3 = 2.15 grams

Mass of O2 = 3.23 grams

Molar mass of NH3 = 17.03 g/mol

Molar mass of O2 = 32 g/mol

volume of N2 produced = 0.550 L

Temperature = 295 K

Pressure = 1.00 atm

Step 2: The balanced equation:

4NH3 (aq) + 3O2 (g) → 2 N2 (g) + 6H2O (l)

Step 3: Calculate moles of NH3

Moles NH3 = Mass NH3 / Molar Mass NH3

Moles NH3 = 2.15 grams / 17.03 g/mol

Moles NH3 = 0.126 moles

Step 4: Calculate moles of O2

Moles O2 = 3.23 grams / 32 g/mol

Moles O2 = 0.101 moles

Step 5: Calculate the limiting reactant

For 4 moles NH3 consumed, we need 3 moles of O2 to produce, 2 moles of N2 and 6 moles of H2O

NH3 is the limiting reactant. It will completely be consumed ( 0.126 moles).

O2 is in excess, there will be 3/4 * 0.126 = 0.0945 moles consumed

There will remain 0.101 - 0.945 = 0.0065 moles of O2

Step 6: Calculate moles of N2

For 4 moles NH3 consumed, we need 3 moles of O2 to produce, 2 moles of N2 and 6 moles of H2O

For 4 moles NH3 , we'll have 2 moles of N2 produced

For 0.126 moles NH3 consumed, we'll have 0.063 moles of N2 produced.

Step 7: Calculate volume of N2 produced

p*V = n*R*T

⇒ with p = the pressure of the gas = 1.00 atm

⇒ with V = the volume = TO BE DETERMINED

⇒ with n = the number of moles N2 = 0.063 moles

⇒ with R = the gasconstant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 295

V = (nRT)/p

V = (0.063*0.08206*295)/1

V = 1.525 L = theoretical yield

Step 8: Calculate the % yield

% yield = actual yield / theoretical yield

% yield = (0.550 L / 1.525 L)*100%

% yield = 36.1 %

4 0

3 years ago

What is a limiting factor?

alukav5142 [94]

A limiting factor helps an organism outcompete other organisms i think that’s the answer

5 0

3 years ago