What is the difference in 3S, 3D and 3P? Is P higher than S in some cases in terms of 3?

Which is the strongest base among NO3-, So4-, CH3COO- and Cl-

vodomira [7]

Answer:

CH3COO-

Explanation:

The conjugate base if a weak acid is a strong base, and the conjugate base of a strong acid is a weak base.

Acetic acid is a weak acid, so its conjugate base (CH3COO-) is a strong base.

Nitric, sulfuric, and hydrochloric acids are all strong acid. Their conjugate bases (NO3-, SO4^2-, and Cl- are all weak bases.

3 0

3 years ago

PH que tendra una solución cuya concentracion de [OH-] = 1x10-4

natta225 [31]

I don’t understand wha u just said but thanks for the points

7 0

3 years ago

Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and is allowed to reach equilibriumdescribed by the equation N2O4(g)↔

Vilka [71]

Answer : The correct option is, (C) 1.1

Solution : Given,

Initial moles of $N_2O_4$ = 1.0 mole

Initial volume of solution = 1.0 L

First we have to calculate the concentration $N_2O_4$ .

$\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}$

$\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M$

The given equilibrium reaction is,

$N_2O_4(g)\rightleftharpoons 2NO_2(g)$

Initially c 0

At equilibrium $(c-c\alpha)$ $2c\alpha$

The expression of $K_c$ will be,

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

$K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}$

where,

$\alpha$ = degree of dissociation = 40 % = 0.4

Now put all the given values in the above expression, we get:

$K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}$

$K_c=\frac{(2\times 1\times 0.4)^2}{(1-1\times 0.4)}$

$K_c=1.066\aprrox 1.1$

Therefore, the value of equilibrium constant for this reaction is, 1.1

4 0

3 years ago

If 40.0 mL of a calcium nitrate solution reacts with excess potassium carbonate to yield 0.524 grams of a precipitate, what is t

tankabanditka [31]

Answer : The molarity of calcium ion on the original solution is, 0.131 M

Explanation :

The balanced chemical reaction is:

$Ca(NO_3)_2+K_2CO_3\rightarrow CaCO_3+3KNO_3$

When calcium nitrate react with potassium carbonate to give calcium carbonate as a precipitate and potassium nitrate.

First we have to calculate the moles of $CaCO_3$

$\text{Moles of }CaCO_3=\frac{\text{Mass of }CaCO_3}{\text{Molar mass of }CaCO_3}$

Given:

Mass of $CaCO_3$ = 0.524 g

Molar mass of $CaCO_3$ = 100 g/mol

$\text{Moles of }CaCO_3=\frac{0.524}{100g/mol}=0.00524mol$

Now we have to calculate the concentration of $CaCO_3$

$\text{Concentration of }CaCO_3=\frac{\text{Moles of }CaCO_3}{\text{Volume of solution}}=\frac{0.00524mol}{0.040L}=0.131M$

Now we have to calculate the concentration of calcium ion.

As, calcium carbonate dissociate to give calcium ion and carbonate ion.

$CaCO_3\rightarrow Ca^{2+}+CO_3^{2-}$

So,

Concentration of calcium ion = Concentration of $CaCO_3$ = 0.131 M

Thus, the concentration or molarity of calcium ion on the original solution is, 0.131 M

4 0

3 years ago

If 2.60 g of NaBr are dissolved in enough water to make 160. mL of solution, what is the molar concentration of

navik [9.2K]

This problem has two parts; the first one asking for the concentration of NaBr given both its mass and volume and the second one asking for its volume given both mass and concentration. The answers turn out to be 0.158 M and 211 mL.

<h3>Molarity</h3>

In chemistry, the use of units of concentration depends on both the substances to analyze and their amounts. In such a way, for molarity, one needs the following relationship between the moles of solute and volume of solution:

$M=\frac{n}{V}$

Thus, for the first part of the problem we first calculate the moles in 2.60 g of NaBr via its molar mass:

$2.60g*\frac{1mol}{102.89g} =0.0253mol$

Next, we convert the 160. mL to L by dividing by 1000 in order to obtain 0.160 L to subsequently calculate the molarity:

$M=\frac{0.0253mol}{0.160L}=0.158M$

Next, since the moles remain the same and for the second part we are asked for the volume given the concentration, one can solve for the volume so as to obtain:

$V=\frac{n}{M} =\frac{0.158M}{0.120mol/L}\\ \\V=0.211L$

That in milliliters turns out to be:

$V=0.211L*\frac{1000mL}{1L}=211mL$

Learn more about molarity: brainly.com/question/10053901

6 0

2 years ago

What is the difference in 3S, 3D and 3P? Is P higher than S in some cases in terms of 3?​

What is the difference in 3S, 3D and 3P? Is P higher than S in some cases in terms of 3?