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ElenaW [278]
3 years ago
13

Calculate the amount of F.A.S required to prepare 1000 ml of 0.1 molar standard solution of F.A.S

Chemistry
1 answer:
RideAnS [48]3 years ago
8 0

Answer:

Preparation and Standardization of 0.1 M Ferrous Ammonium...

Dissolve 40 g of ferrous ammonium sulfate in a previously cooled mixture of 40 ml of sulphuric acid and 200 ml of water.

Dilute with sufficient freshly boiled and cooled water to produce 1000 ml.

Standardize the solution in the following manner.

Explanation:

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HNO3 + H2O → H3O^+ + NO3<br> which ones are acids and which ones are bases
nlexa [21]

Answer:

  • HNO₃ and  H₃O⁺ are acids
  • H₂O and  NO₃⁻ are bases

Explanation:

The chemical equation is:

  • HNO₃ + H₂O → H₃O⁺ + NO₃⁻

There are several definitions of acid and bases: Arrhenius', Bronsted-Lowry's and Lewis'.

Bronsted-Lowry model defines and <em>acid</em> as a donor of protons, H⁺.

In the given equation HNO₃ is such substance: it releases an donates its hdyrogen to form the H₃O⁺ ion.

On the other hand, a <em>base</em> is a substance that accepts protons.

In the reaction shown, H₂O accepts the proton from HNO₃ to form H₃O⁺.

Thus, H₂O is a base.

In turn, on the reactant sides the substances can be classified as acids or bases.

H₃O⁺ contain an hydrogen that can be donated and form H₂O; thus, it is an acid (the conjugated acid), and NO₃⁻ can accept a proton to form HNO₃; thus it is a base (the conjugated base).

4 0
3 years ago
g A solution contains 100mM NaCl, 20mM CaCl2, and 20mM urea. We would say this solution is __________ compared to a 300 mOsM sol
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Answer:

A solution contains 100mM NaCl, 20mM CaCl2, and 20mM urea. We would say this solution is hypotonic compared to a 300 mOsM solution and hypotonic compared to a cell with 300 mOsM (non-penetrating solutes) interior.

Explanation:

The osmolarity is calculated from the molar concentration of the active particles in the solution. We have a solution that is composed of NaCl, CaCl₂ and urea.

When they are dissolved in water, they dissociate into particles as follows:

NaCl → Na⁺ + Cl⁻  (2 particles per compound)

CaCl₂ → Ca²⁺ + 2 Cl⁻ (3 particles per compound)

urea: not dissociation (1 particle per compound)

Then, we have to calculate the osmolarity of the solution. We multiply the molarity of each compound by the number of particles produced by the compound in water:

Osm = (100 mM NaCl x 2) + (20 mM CaCl₂ x 3) + (20 mM urea x 1) = 280 mOsm

Compared with 300 mOsm, 280 mOsm has a lower osmolarity, so it is a hypotonic solution.

To compare with a cell's osmolarity, we have to consider only the non-penetrating solutes. Urea is considered a penetrating solute for mammalian cells. So, the osmolarity of non-penetrating solutes (NaCl  and CaCl₂) is calculated as:

Osm (non-penetrating solutes) = (100 mM NaCl x 2) + (20 mM CaCl₂ x 3) = 260 mOsm

Therefore, we have:

Compared to 300 mOsm solution ⇒ 280 mOsm solution is a hypotonic solution

Compared to a cell with 300 mOsm ⇒ 260 mOsm solution is hypotonic

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