WHaty are the Chemical properties of Oxygen?

The molar mass of a solid carboxylic acid is determined by titrating a known mass of the acid with a standardized solution of Na

laiz [17]

The answer is: II.The endpoint is recorded when the solution is dark red in color rather than light pink.

The endpoint is the point at which the indicator changes colour in a colourimetric titration and that is point when titration must stop.

Phenolphthalein is colorless in acidic solutions and pink in basic solutions. If this indicator change color to dark red, more base is added and endpoint is not accurate.

If the the acid is spilled before titration, that does not make endpoint wrong and molar mass can be calculated.

In this example we can take acetic acid as carboxylic acid; basic salt sodium acetate CH₃COONa is formed from the reaction between weak acid (in this example acetic acid CH₃COOH) and strong base (in this example sodium acetate NaOH).

Balanced chemical reaction of acetic acid and sodium hydroxide:

CH₃COOH(aq) + NaOH(aq) → CH₃COONa(aq) + H₂O(l).

Neutralization is is reaction in which an acid (in this example vinegar or acetic acid CH₃COOH) and a base react quantitatively with each other.

8 0

3 years ago

You have a red ballooe that has a volume of 20 liters, a pressure of 1.5 atmospheres and a temperature of 28 C. What is the volu

ki77a [65]

Explanation:

The given data is as follows.

$P_{1}$ = 1.5 atm, $V_{1}$ = 20 L, $T_{1}$ = (28 + 273) K = 301 K

$P_{2}$ = 5 atm, $V_{2}$ = ?, $T_{2}$ = (50 + 273) K = 323 K

Formula to calculate the volume will be as follows.

$\frac{P_{1} \times V_{1}}{T_{1}}$ = $\frac{P_{2} \times V_{2}}{T_{2}}$

Putting the given values into the above formula as follows.

$\frac{P_{1} \times V_{1}}{T_{1}}$ = $\frac{P_{2} \times V_{2}}{T_{2}}$

$\frac{1.5 atm \times 20 L}{301 K}$ = $\frac{5 atm \times V_{2}}{323 K}$

$V_{2}$ = 0.64 L

Thus, we can conclude that the change in volume of the balloon will be 0.64 L.

6 0

3 years ago

Why are covalent bonds between hydrogen and nitrogen or oxygen polar? see section 2.1 ( page 57) ?

ASHA 777 [7]

Polar molecules exhibit an unequal balance of charges between the individual elements of the compound. This is brought about by the large difference in their electronegativities. The H atom has the least amount of electronegativity. Then, it is a known periodic trend, that as you go downwards in a group, electronegativity decreases, and increase as you go from left to right. Thus, you can deduce that the most electronegative elements are found in the upper right corner which includes O, N and F atoms. Any bond created between Hydrogen and any of O, N and F atoms is a polar bond.

3 0

3 years ago

This chemical reaction is unbalanced: ___Pb(NO₃)₂(aq) + ___Li₂SO₄(aq) → ____PbSO₄(s) + ____LiNO₃(aq) Which coefficient should ap

kakasveta [241]

Answer: 2

Explanation:

4 0

3 years ago

First-order reaction that results in the destruction of a pollutant has a rate constant of 0.l/day. (a) how many days will it ta

Klio2033 [76]

Rate equation for first order reaction is as follows:

$t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}$

Here, k is rate constant of the reaction, t is time of the reaction, $A_{0}$ is initial concentration and $A_{t}$ is concentration at time t.

The rate constant of the reaction is $0.1 day^{-1}$ .

(a) Let the initial concentration be 100, If 90% of the chemical is destroyed, the chemical present at time t will be 100-90=10, on putting the values,

$t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{10}=23.03 days$

Thus, time required to destroy 90% of the chemical is 23.03 days.

(b) Let the initial concentration be 100, If 99% of the chemical is destroyed, the chemical present at time t will be 100-99=1, on putting the values,

$t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{1}=46.06 days$

Thus, time required to destroy 99% of the chemical is 46.06 days.

(c) Let the initial concentration be 100, If 99.9% of the chemical is destroyed, the chemical present at time t will be 100-99.9=0.1, on putting the values,

$t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{0.1}=69.09 days$

Thus, time required to destroy 99.9% of the chemical is 69.09 days.

5 0

3 years ago