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Lady_Fox [76]
3 years ago
15

Pls pls I need help to solve this questions

Chemistry
1 answer:
cupoosta [38]3 years ago
6 0
1) • Solubility is the ability of a substance to be dissolved especially in water
• Unit; mg/L (milligrams per liter)

2) A saturated solution is a chemical solution containing the maximum concentration of a solute dissolved in the solvent.

3) To detoxify any acid that was previously used in it

4) CAN’T SEE THE MAIN QUESTIONS

5) CAN’T SEE THE MAIN QUESTIONS
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1. Ethanol and 2,6-dimethylphenol both have hydroxyl groups that can be deprotonated by a base. Why does ethanol act as a solven
Volgvan

Anions are solvated in protic hydrogen-bonding solvents (such as ethanol). Consequently, nucleophiles are less reactive. Since soft nucleophiles are less strongly solvated than hard nucleophiles, these solvents boost the relative reactivity of soft anions.

<h3>Ethanol is either a nucleophile or a base.</h3>

The ethanol is a base  Because carbocation is an extremely reactive species, a base or nucleophile as weak as ethanol can replace or remove it. SN1 and E1 would not be conceivable without the carbocation or a strong departing group.

<h3>How do solvents impact anionic nucleophile's reactivity?</h3>

In polar aprotic solvents, nucleophilic substitution reactions of anionic nucleophiles often proceed more quickly. The normal relative reactivity order in such solvents (like DMSO)is Anions are solvated in protic hydrogen-bonding solvents (such as ethanol). Consequently, nucleophiles are less reactive.

Learn more about nucleophiles here:-

brainly.com/question/27127109

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3 0
1 year ago
How is the concentration of h3o ions and oh- ions different in an acid solution and a basic soulution?
Lubov Fominskaja [6]
In an acidic solution, the concentration of H+ is greater than the concentration of OH-. The pH will be less than 7.
In a basic solution, the concentration of OH- is greater than the concentration of H+. The pH will be greater than 7.
In a neutral solution, the concentration of H+ ions to OH-ions will be equal, and will therefore have a pH of 7. (This is due to water autoionization, which we usually ignore because it is small in other circumstances.)
5 0
3 years ago
A balloon occupies a volume of 2.00 l at 40.0oc. how much volume will it occupy at 30.0oc?
hodyreva [135]
<span>Charles' law says "at a constant pressure, the volume of a fixed amount of gas is directly proportional to its absolute temperature".

V </span>α T

Where V is the volume and T is the temperature in Kelvin of the gas. We can use this for two situations as,
V₁/T₁ = V₂/T₂

V₁ = 2.00 L
T₁ = 40.0 ⁰C = 313 K
V₂ = ?
T₂ = 30.0 ⁰C = 303 K

By applying the formula,
2.00 L / 313 K = V₂ / 303 K                   
                   V₂ = (2.00 L / 313 K) x 303 K
                   V₂ = 1.94 L

Hence, the volume of the balloon at 30.0 ⁰C  is 1.94 L
7 0
3 years ago
If the volume and Kelvin temperature are reduced by one-half what will happen to the pressure???
jolli1 [7]

Answer:

the pressure will decrease

Explanation:

6 0
3 years ago
In Lab 10 you make a stock solution of salicylic acid, and then four dilutions. The stock solution is made by diluting 5.00 ml o
Zigmanuir [339]

Answer:

Stock  solution =  1.25 *10^-3 M

Dilution 1 = 5*10^-4 M

Dilution 2= 3.75 * 10^-4 M

Dilution 3 = 2.5 *10^-4 M

Dilution 4 = 1.25 *10^-4 M

Explanation:

<u>Step 1:</u> Data given

The stock solution is made by diluting 5.00 ml of 1.250 x 10-2 M salicylic acid in 50.00 mL of solution.

<u>Step 2</u>: Calculate the concentration of the stock solution:

M1*V1 = M2*V2

⇒ with M1 = the initial concentration = 1.250 *10^-2 M

⇒ with V1 = 5 mL = 5*10^-3 L

⇒ with M2 = TO BE DETERMINED

⇒ with V2 = 50 mL = 50 *10^-3 L

M2 = (M1*V1)/V2

M2 = (1.250 *10^-2 * 5*10^-3 L) / 50 *10^-3

M2 = 0.00125 M = 1.25 *10^-3 M

<u>Step 3:</u> Calculate dilution 1

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 10 *10^-3)/(25*10^-3L)

M2 = 0.0005 M = 5*10^-4 M

<u>Step 4</u>: Calculate dilution 2

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 10 * 7.5*10^-3)/(25*10^-3)

M2 = 0.000375 M = 3.75 * 10^-4 M

<u>Step 5:</u> Calculate dilution 3

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 5*10^-3) /(25*10^-3)

M2 = 0.00025 M = 2.5 *10^-4 M

<u>Step 6</u>: Calculate dilution 4

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 2.5*10^-3)/(25*10^-3)

M2 = 0.000125 M = 1.25 *10^-4 M

5 0
3 years ago
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