Sickle cell is anemia is a deadly,recessive disease.How is sickle cell anemia inherited

What can be said about a small Ksp value?

fredd [130]

Answer:

The correct answer is option B.

Explanation:

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.

The equation for the ionization of the magnesium hydroxide is given as:

$A_xB_y\leftrightharpoons xA^{y+}+yB^{x-}$

xS yS

Here, S = concentration of compound dissolved in solvent via dissociation

Expression for the solubility product of $A_xB_y$ will be:

$K_{sp}=[A^{y+}]^x[B^{x-}]^y\\\\K_{sp}=(xS)^x\times (yS)^y$

As we can see,

$K_{sp}\propto [S]$

Higher the value of [S] more will be solubility of compound in a solvent.Hence, more soluble it will be.
Lower the value of [S] less will be solubility of compound in a solvent.Hence, less soluble it will be.

Hence, the correct answer is option B.

7 0

3 years ago

What is the weakness of particle theory?

stich3 [128]

Answer:

The particle model does not take into account: the size and shape of particles. the space between particles.

5 0

3 years ago

Write a story of your life when you were hurted by someone whom you trusted blindly...

SOVA2 [1]

Answer:

Sis I think it happened with me but I am not able to remember if u want u can share if it happened with u

4 0

3 years ago

Using 18.8L, calculate the volume (in L) of water vapour that should be produced by the reaction of 35.0 g

victus00 [196]

Answer:

A. 18.8L

B. 75.2L of H2O.

Explanation:

A. Determination of the volume of 35g of C3H8.

Date obtained from the question include the following:

Mass of C3H8 = 35g

Temperature (T) = 40°C

Pressure (P) = 110KPa

Volume (V) =..?

Next, we shall determine the number of mole (n) in 35g of C3H8. This is illustrated below:

Molar mass of C3H8 = (3x12) + (8x1) = 44g/mol

Mass of C3H8 = 35g

Mole of C3H8 =..?

Mole = mass /molar mass

Mole of C3H8 = 35/44

Mole of C3H8 = 0.795 mole

Finally, we shall determine the volume of 35g of C3H8 as follow:

Temperature (T) = 40°C = 40°C + 273 = 313K

Pressure (P) = 110KPa

Number of mole (n) = 0.795 mole

Gas constant (R) = 8.314 KPa.L/Kmol

Volume (V) =..?

PV = nRT

110 x V = 0.795 x 8.314 x 313

Divide both side by 110

V = (0.795 x 8.314 x 313)/110

V = 18.8L

Therefore, the volume of 35g of C3H8 under the conditions given is 18.8L

B. Determination of the volume of water vapour produced by the reaction of 35g of propane, C3H8.

From the calculations made in (A) above, 35g of C3H8 is equivalent to 18.8L of C3H8.

Thus, we can obtain the volume of water vapour produced as follow:

C3H8 + 5O2 —> 3CO2 + 4H2O

From the balanced equation above,

1L of C3H8 reacted to produce 4L of H2O.

Therefore, 18.8L of C3H8 will react to produce = (18.8 x 4)/1 = 75.2L of H2O.

Therefore, 75.2L of H2O were produced from the reaction.

4 0

3 years ago

The reaction 2NO(g)+O2(g)−→−2NO2(g) is second order in NO and first order in O2. When [NO]=0.040M, and [O2]=0.035M, the observed

Oksanka [162]

Answer:

(a) The rate of disappearance of $O_{2}$ is: $4.65*10^{-5}$ M/s

(b) The value of rate constant is: 0.83036 $M^{-2}s^{-1}$

(c) The units of rate constant is: $M^{-2}s^{-1}$

(d) The rate will increase by a factor of 3.24

Explanation:

The rate of a reaction can be expressed in terms of the concentrations of the reactants and products in accordance with the balanced equation.

For the given reaction:

$2NO(g)+O_{2}->2NO_{2}$

rate = $-\frac{1}{2} \frac{d}{dt}[NO]$ = $-\frac{d}{dt}[O_{2}]$ = $\frac{1}{2}\frac{d}{dt}[NO_{2}]$ -----(1)

According to the question, the reaction is second order in NO and first order in $O_{2}$ .

Then we can say that, rate = k $[NO]^{2}[O_{2}]$ -----(2)

where k is the rate constant.

The rate of disappearance of NO is given:

$-\frac{d}{dt}[NO]$ = $9.3*10^{-5}$ M/s.

(a) From (1), we can get the rate of disappearance of $O_{2}$ .

Rate of disappearance of $O_{2}$ = $-\frac{d}{dt}[O_{2}]$ = (0.5)*( $9.3*10^{-5}$ ) M/s = $4.65*10^{-5}$ M/s.

(b) The rate of the reaction can be obtained from (1).

rate = $-\frac{1}{2} \frac{d}{dt}[NO]$ = (0.5)*( $9.3*10^{-5}$ )

rate = $4.65*10^{-5}$ M/s

The value of rate constant can be obtained by using (2).

rate constant = k = $\frac{rate}{[NO]^{2}[O_{2}]}$

k = $\frac{4.65*10^{-5}}{(0.040)^{2}(0.035)}$ = 0.83036 $M^{-2}s^{-1}$

(c) The units of the rate constant can be obtained from (2).

k = $\frac{rate}{[NO]^{2}[O_{2}]}$

Substituting the units of rate as M/s and concentrations as M, we get:

$\frac{Ms^{-1} }{M^{3}}$ = $M^{-2}s^{-1}$

(d) The reaction is second order in NO. Rate is proportional to square of the concentration of NO.

$rate\alpha [NO]^{2}$

If the concentration of NO increases by a factor of 1.8, the rate will increase by a factor of $(1.8)^{2}$ = 3.24

5 0

3 years ago