The photoelectric effect is the basis for how solar energy panels work today.

The bus will stop here into question tag

Vikentia [17]

Answer:

Huh????????????????????

7 0

3 years ago

What is the ph of a buffer that consists of 0.45 m ch3cooh and 0.35 m ch3coona? ka = 1.8 × 10–5?

oee [108]

Answer is: pH of a buffer is 4.64.

ck(CH₃COOH) = 0.45 M.

cs(CH₃COONa) = 0.35 M.

Ka = 1.8·10⁻⁵.

pKa = -logKa.
pKa = -log(1.8·10⁻⁵) = 4.75.
Henderson–Hasselbalch equation: pH = pKa + log(cs/ck).
pH = 4.75 + log(0.35M/0.45M).
pH = 4.75 - 0.11.
pH = 4.64.

pH (potential of hydrogen) is a numeric scale used to specify the acidity or basicity an aqueous solution.

8 0

4 years ago

ZnS has two common polymorphs: Sphalerite and Wurtzite. Based on the analysis of Medelung constants alone, predict which polymor

denis-greek [22]

Answer:

Wurzite has a higher Madelung constant when compared with Sphalerite therefore Wurizite is more stable.

Explanation:

Zinc Sulphide is a chemical compound that exhibits polymorphism. This means Zinc Sulphide can exists in different forms as a compound. Zinc Sulphide has two common polymorphs named; Wurtzite and Sphalerite .

This two polymorphs are crystalline in structure. Wurtzite is hexagonal in shape while Sphalerite is cubic in shape.

Madelung constant was named after German Physicist Erwin Madelung. Madelung constant is dependent on the the way ions are arranged in a solid substance or molecule. It is used to calculate the amount of energy required to move an ion from one point to the another in a crystal substance.

Madelung constant for Wurtzite is 1.64132 and for Sphalerite is 1.63806. Due to the fact that Wurtzite has a slightly higher Madelung constant that Sphalerite, it tends to be more stable than Sphalerite.

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3 0

3 years ago

Benzoyl peroxide, the substance most widely used against acne, has a half-life of 9.8 × 103 days when refrigerated. How long wil

prohojiy [21]

Answer:

$7.3\times 10^2\ days$

Explanation:

Given that:

Half life = $9.8\times 10^3$ days

$t_{1/2}=\frac{\ln2}{k}$

Where, k is rate constant

So,

$k=\frac{\ln2}{t_{1/2}}$

$k=\frac{\ln2}{9.8\times 10^3}\ days^{-1}$

The rate constant, k = 0.00007 days⁻¹

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given:

5 % is lost which means that 0.05 of $[A_0]$ is decomposed. So,

$\frac {[A_t]}{[A_0]}$ = 1 - 0.05 = 0.95

t = ?

$\frac {[A_t]}{[A_0]}=e^{-k\times t}$

$0.95=e^{-0.00007\times t}$

t = 732.76 days = $7.3\times 10^2\ days$

7 0

3 years ago

What is the orbital diagram for phosphorus

scoray [572]

The orbital diagram for phosphorus consists of five electrons in the third shell, eight in the second and two in the first shell, closest to the nucleus. The atomic number of phosphorus is 15. This number indicates the total number of electrons.

Ta Da!!!

Just do a little research!!!

Or not you know, that's what we're here for...

4 0

3 years ago