Which organism has cerebral ganglia in its head connected to ladder-like arrangements of nerve fibers?

The stock concentration of dye is 3.4E-5M. The stock concentration of bleach is 0.36M. Assuming that the reaction goes to comple

Aleks04 [339]

Answer:

0.036 M

Explanation:

To do this, let's mark the dye as D and bleach as B.

We have the concentrations of both, and we already know that they react in a 1:1 mole ratio. The total volume of reaction is 9 + 1 = 10 mL or 0.010 L, and we hava both concentrations.

The problem already states that the dye reacts completely, so this is the limiting reagent, while bleach is the excess.

To know the remaining amount of bleach, we need to do this with the moles. First, let's calculate the initial moles of D and B:

moles D = 3.4x10⁻⁵ * 0.009 = 3.06x10⁻⁷ moles

moles B = 0.36 * 0.001 = 3.6x10⁻⁴ moles

Now that we have the moles, and that we know that all the dye reacts completely, let's see how many moles of bleach are left:

moles of B remaining = 3.6x10⁻⁴ - 3.06x10⁻⁷ = 3.597x10⁻⁴ moles

These are the moles presents of B after the reaction has been made. The concentration of the same will be:

[B] = 3.597x10⁻⁴ / 0.010

[B] = 0.0357

With 2 SF it would be:

[B] = 3.6x10⁻² M

5 0

1 year ago

Consider a general reaction A ( aq ) enzyme ⇌ B ( aq ) A(aq)⇌enzymeB(aq) The Δ G ° ′ ΔG°′ of the reaction is − 5.980 kJ ⋅ mol −

Grace [21]

Answer : The value of $K_{eq}$ is, 11.2

The value of $\Delta G_{rxn}$ is -9.04 kJ/mol

Explanation :

The relation between the equilibrium constant and standard Gibbs free energy is:

$\Delta G^o=-RT\times \ln K_{eq}$

where,

$\Delta G^o$ = standard Gibbs free energy = -5.980 kJ/mol = -5980 J/mol

R = gas constant = 8.314 J/K.mol

T = temperature = $25^oC=273+25=298K$

$K_{eq}$ = equilibrium constant = ?

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

$\Delta G^o=-RT\times \ln K_{eq}$

$-5980J/mol=-(8.314J/K.mol)\times (298K)\times \ln K_{eq}$

$K_{eq}=11.2$

Thus, the value of $K_{eq}$ is, 11.2

Now we have to calculate the $\Delta G_{rxn}$ .

The formula used for $\Delta G_{rxn}$ is:

The given reaction is:

$A(aq)\rightleftharpoons B(aq)$

$\Delta G_{rxn}=\Delta G^o+RT\ln Q$

$\Delta G_{rxn}=\Delta G^o+RT\ln \frac{[B]}{[A]}$ ............(1)

where,

$\Delta G_{rxn}$ = Gibbs free energy for the reaction = ?

$\Delta G_^o$ = standard Gibbs free energy = -30.5 kJ/mol

R = gas constant = $8.314\times 10^{-3}kJ/mole.K$

T = temperature = $37.0^oC=273+37.0=310K$

Q = reaction quotient

[A] = concentration of A = 1.8 M

[B] = concentration of B = 0.55 M

Now put all the given values in the above formula 1, we get:

$\Delta G_{rxn}=(-5980J/mol)+[(8.314J/mole.K)\times (310K)\times \ln (\frac{0.55}{1.8})$

$\Delta G_{rxn}=-9035.75J/mol=-9.04kJ/mol$

Therefore, the value of $\Delta G_{rxn}$ is -9.04 kJ/mol

3 0

1 year ago

A tank of oxygen has a volume of 40.0L and is held at a pressure of 159atm at 25∘C. What volume of O2 gas (in liters) would ther

Advocard [28]

Answer:

V2 = 6616 L

Explanation:

From the question;

Initial volume = 40L

Initial Pressure, P1 = 159atm

Initial Temperature T1 = 25 + 273 = 298K (Upon converting to Kelvin unit)

Final Volume, V2 = ?

Final Pressure, P2 = 1 atm

Final Temperature T2 = 37 + 273= 310K (Upon converting to Kelvin unit)

These quantities are related by the equation;

P1V1 / T1 = P2V2 / T2

V2 = T2 * P1 * V1 / T1 * P2

V2 = 310 * 159 * 40 / (298 * 1)

V2 = 6616 L

6 0

1 year ago

A 1.0-gram sample of solid iodine is placed in a tube and the tube is sealed after all of the air is removed. The tube and the s

yKpoI14uk [10]

Answer:

27.0

Explanation:

Because Mass can neither be created nor be destroyed hence total mass of sample of iodine and tube remain equal as it is sealed.

7 0

1 year ago

How many liters of a 0.0550 M NaF solution contain 0.163 moles of NaF?

Kobotan [32]

Answer : The volume of solution will be 2.96 liters.

Explanation :

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

$\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

In this question, the solute is NaF.

Now put all the given values in this formula, we get:

$0.0550M=\frac{0.163mole}{\text{Volume of solution (in L)}}$

$\text{Volume of solution (in L)}=\frac{0.163mole}{0.0550M}$

$\text{Volume of solution (in L)}=2.96L$

Therefore, the volume of solution will be 2.96 liters.

3 0

1 year ago