How many milligrams of MgI2 must be added to 257.7 mL of 0.087 M KI to produce a solution with [I−] = 0.1000 M?

The formula for Molarity is given by:

                                  M = moles / V
To isolate V,
                              V = moles / M ------------------(1)
Moles can also be calculated as:
                                  moles = mass / M.mass -------------(2)
Substituting the value of moles from equation 2 into equation 1 yields:
                                  V = (mass / M.mass) / M
Plugging in the numbers gives:
                                  V = (45 g / 164 g/mol) / 1.3 mol/dm³

                                       V = 0.21 dm³.

Answer:

The work done in this process will be considered Negative.

Explanation:

The energy transferred by the system to the environment is negative

Therefore, if work is done on the system, it is labeled as positive. Conversely, when work is done by the system, it is regarded as negative.

In this scenario, the argon gas is expanding, and the work is exerted by the system into the surroundings (container), making the sign Negative.

Thus, the result for the work pertaining to this process will carry a Negative sign.

Answer:

The rate law for the decomposition reaction is:

The unit for the rate constant will be

Explanation:

The rate law can be expressed as:

..[1]

When the drug concentration is tripled, the decomposition rate rises by a factor of nine.

...[2]

[1] ÷ [2]

Solving for x results in:

x = 2.

This indicates a second-order reaction.

The decomposition reaction's rate law is:

The unit for the rate constant will be:

The unit for the rate constant will be .

Answer:

a. H₂O (conjugate acid); b. OH⁻ (conjugate base), H₃O⁺ (conjugate acid); c. H₂CO₃ (conjugate acid), CO₃⁻² (conjugate base); d. NH₄⁺ (conjugate strong acid) e. H₂SO₄ (conjugate acid), SO₄⁻² (conjugate base); f. No conjugate acid or base exists; g. H₂S (conjugate acid), S⁻² (conjugate base);

h. H₄N₂ (conjugate base)

Explanation:

a. OH⁻ + H⁺ ⇄ H₂O

The hydroxide functions as a Bronsted-Lowry base, allowing it to capture a proton, thus water serves as the conjugate acid.

b. H₂O is amphoteric, capable of acting as either an acid or a base. As a base, its conjugate acid is H₃O⁺, whereas as an acid, its conjugate base is OH⁻.

c. HCO₃⁻ + H⁺ ⇄ H₂CO₃

HCO₃⁻ + H₂O ⇄ CO₃⁻² + H₃O⁺

Bicarbonate is also amphoteric. When it captures a proton, it forms carbonic acid as the conjugate acid when acting as a base. When HCO₃⁻ acts as an acid and releases a proton, carbonate becomes the conjugate base.

d. Ammonia functions as a weak base, with ammonium being the conjugate strong acid.

NH₃ + H₂O ⇄ NH₄⁺ + OH⁻

e. Another amphoteric compound. Acid sulfate can function as both an acid and a base.

(similar to bicarbonate). Acting as a base yields sulfuric acid as the conjugate acid, while acting as an acid leads to sulfate as the conjugate base.

HSO₄⁻ + H₂O ⇄ SO₄⁻² + H₃O⁺

HSO₄⁻ + H⁺ ⇄ H₂SO₄

f. H₂O₂ does not accept H⁺ or OH⁻ nor does it expel H⁺. It’s neutral and does not function as an acid or base.

g. HS⁻ is amphoteric.

HS⁻ + H⁺ ⇄ H₂S

HS⁻ + H₂O ⇄ S⁻² + H₃O⁺

This is similar to the case of bicarbonate or acid sulfate.

h. H₅N₂⁺ + H₂O ⇄ H₄N₂ + H₃O⁺

Hydrazinium acts as an acid, making hydrazine its conjugate base.

Response: The rate constant at 525 K is,

Rationale:

Based on the Arrhenius equation,

or,

where,

= rate constant when =

= rate constant when =?

= activation energy for the process =

R = gas constant = 8.314 J/mole.K

= initial temperature = 701 K

= final temperature = 525 K

Substituting the provided values into this formula yields:

Thus, the rate constant at 525 K is,