What is the y and x axis?

A 3.96x10^-24 M solution of compound A exhibited an absorbance of 0.624 at 238 nm in a 1.000-cm cuvet; a blank solution containi

viktelen [127]

Actual question from source:-

A 3.96x10-4 M solution of compound A exhibited an absorbance of 0.624 at 238 nm in a 1.000 cm cuvette. A blank had an absorbance of 0.029. The absorbance of an unknown solution of compound A was 0.375. Find the concentration of A in the unknown.

Answer:

Molar absorptivity of compound A = $1502.53\ {Ms}^{-1}$

Explanation:

According to the Lambert's Beer law:-

$A=\epsilon l c$

Where, A is the absorbance

l is the path length

$\epsilon$ is the molar absorptivity

c is the concentration.

Given that:-

c = $3.96\times 10^{-4}\ M$

Path length = 1.000 cm

Absorbance observed = 0.624

Absorbance blank = 0.029

A = 0.624 - 0.029 = 0.595

So, applying the values in the Lambert Beer's law as shown below:-

$0.595=\epsilon\times 1.000\ cm\times 3.96\times 10^{-4}\ M$

$\epsilon=\frac{0.595}{3.96\times 10^{-4}}\ {Ms}^{-1}=1502.53\ {Ms}^{-1}$

<u>Molar absorptivity of compound A = $1502.53\ {Ms}^{-1}$ </u>

4 0

4 years ago

Solve 123,000 m x 3,234 m = ?

Leya [2.2K]

Answer:

397782000 m²

Explanation:

Given data:

123000 m × 3234 m =?

Solution:

123000 m × 3234 m = 397782000 m²

when we multiply the given quantities having same unit i.e meter the answer should be m² . It could be the area, and its unit is m².

Area = length (m) × width (m)

3 0

3 years ago

A glucose solution is administered intravenously into the bloodstream at a constant rate r. As the glucose is added, it is conve

Helen [10]

Answer:

$C(t)=\frac{r}{k}-(\frac{r-kC_o}{k})e^{-kt}$

Explanation:

The differential equation is given as:

$\frac{dC}{dt} = r- kC$

$\frac{dC}{r- kC} = dt$

Taking integral of both sides; we have:

$\int\limits \frac{dC}{r- kC} = \int\limits dt$

$-\frac{1}{k} In(r-kC) = t+D\\In(r-kC)=-kt-kD$

$r-kC=e^{-kt-kD}$

$r-kC=e^{-kt}e^{-kD}$

$r-kC=Ae^{-kt}$

$kC=r-Ae^{-kt}$

$C=\frac{r}{k}-\frac{A}{k}e^{-kt}$

$C(t)=\frac{r}{k}-\frac{A}{k}e^{-kt}$ ------- equation (1)

If C(0)= $C_o$ ; we have:

C(0)= $\frac{r}{k}-\frac{A}{k}e^0$ (where; A is an integration constant)

$C_o = \frac{r}{k}- \frac{A}{k}$

$C_o=\frac{r-A}{k}$

$kC_o=r-A$

$A=r-kC_o$

Substituting $A=r-kC_o$ into equation (1); we have;

$C(t)=\frac{r}{k}-(\frac{r-kC_o}{k})e^{-kt}$

3 0

4 years ago

13. Describe the flow of thermal energy.<br> Plz help

just olya [345]

Answer: Thermal energy refers to the energy contained within a system that is responsible for its temperature. Heat is the flow of thermal energy. ... This energy is usually in the form of low-level thermal energy. Here, low-level means that the temperature associated with the thermal energy is close to that of the environment. hope this helps. Can u pls give me brainliest

Explanation:

4 0

3 years ago

In another experiment, a 0.150 M BF4^-(aq) solution is prepared by dissolving NaBF4(s) in distilled water. The BF4^-(aq) ions in

Ilia_Sergeevich [38]

Answer:

A) Forward rate = 1.1934 × 10^(-4) M/min

B) I disagree with the claim

Explanation:

A) We are told that [HF] reaches a constant value of 0.0174 M at equilibrium.

The reversible reaction given to us is;

BF4-(aq) +H20(l) → BF3OH-(aq) + HF(aq)

From this, we can see that the stoichiometric ratio is 1:1:1:1

Thus, concentration of [BF4-] is now;

[BF4-] = 0.150 - 0.0174

[BF4-] = 0.1326 M

From the rate law, we are told the forward rate is kf [BF4-].

We are given Kf = 9.00 × 10^(-4) /min

Thus;

Forward rate = 9.00 × 10^(-4) /min × (0.1326M)

Forward rate = 1.1934 × 10^(-4) M/min

(B) The student claims that the initial rate of the reverse reaction is equal to zero can't be true because at equilibrium, rates for the forward and reverse reactions are usually equal.

Thus, I disagree with the claim.

3 0

3 years ago

What is the y and x axis?​

What is the y and x axis?