CHEMISTRY COLLEGE

Many calculators use photocells to provide their energy. Find the maximum wavelength needed to remove an electron from silver (Φ = 7.59 x 10⁻¹⁹ J). Is silver a good choice for a photocell that uses visible light?

Answers

Answer 1

Answer:

Answer:

$\lambda=2.61* 10^(-9)\ m$ = 261 nm

Silver is not a good choice.

Explanation:

$E=\frac {h* c}{\lambda}$

Where,

h is Plank's constant having value $6.626* 10^(-34)\ Js$

c is the speed of light having value $3* 10^8\ m/s$

$\lambda$ is the wavelength of the light

Given that:- Energy = $7.59* 10^(-19)\ J$

$7.59* 10^(-19)=(6.626* 10^(-34)* 3* 10^8)/(\lambda)$

$7.59* \:10^(26)* \lambda=1.99* 10^(20)$

$\lambda=2.61* 10^(-9)\ m$ = 261 nm

Visible range has a spectrum of 380 to 740 nm

So, Silver is not a good choice.

Answer 2

Answer:

Final answer:

The maximum wavelength needed to remove an electron from silver is approximately 262 nm. Silver is not a good choice for a photocell that uses visible light.

Explanation:

To find the maximum wavelength needed to remove an electron from silver, we can use the work function of silver, which is Φ = 4.73 eV. The threshold wavelength for observing the photoelectric effect in silver can be calculated using Equation 6.16, which is λ = hc/Φ. Substituting the given values, we have λ = (1240 eV⋅nm) / (4.73 eV), which gives us a threshold wavelength of approximately 262 nm. Since visible light ranges from 400 to 700 nm, silver is not a good choice for a photocell that uses visible light.

Learn more about photoelectric effect in silver here:

brainly.com/question/35875610

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On another planet, the isotopes of titanium have the following natural abundances. a. Isotope 46Ti Abundance 70.900% Mass(amu) 45.95263
b. Isotope 48Ti Abundance 10.000% Mass(amu) 47.94795
c. Isotope 50Ti Abundance 19.100% Mass(amu) 49.94479
d. What is the average atomic mass of titanium on that planet?
e. I got 46.9 amu but it is wrong.

Answers

Answer:

Average atomic mass = 46.91466 amu

Explanation:

Step 1: Data given

Isotopes of titanium

46Ti = 70.900% ⇒ 45.95263 amu

48Ti = 10.000 % ⇒ 47.94795 amu

50Ti = 19.100 % ⇒ 49.94479 amu

Step 2: Calculate the average atomic mass of titanium

Average atomic mass = 0.7090 * 45.95263 + 0.10 * 47.94795 + 0.1910 * 49.94479

Average atomic mass = 46.91466 amu

What is the mass of ammonium nitrate in 250 mL of a 75% by mass solution (density = 1.725 g/mL)?

Answers

Explanation:

Density is the mass present in per unit volume.

Mathematically, Density = $(mass)/(volume)$

Therefore, first calculate the mass of solution as follows.

Density = $(mass)/(volume)$

1.725 g/mL = $(mass)/(250 mL)$

mass = 431.25 g

Now, calculate mass of ammonium nitrate as follows.

Percentage by mass = $\frac{\text{mass of ammonium nitrate}}{\text{mass of solution}} * 100$

75 = $\frac{\text{mass of ammonium nitrate}}{431.25g} * 100$

Mass of ammonium nitrate = 323.43 g

Thus, we can conclude that mass of ammonium nitrate is 323.43 g.

1) We need to find mass of the solution

m=D*V
D= 1.725 g/mL
V= 250 mL
m=1.725 g/mL*250 mL= 431.25 g

2) 75% = 0.75
0.75*431.25 ≈ 323 g of NH4NO3

Nf3 Rotate the molecule until you have a feeling for its three-dimensional shape. How many atoms are bonded to the central atom?

Answers

Answer:

Three atoms are attached to the central atom in NF3.

Explanation:

The central atom is always regarded as the atom having the least electronegativity in a molecule or ion. We can decide on what atom should be the central atom by comparing the relative electro negativities of the atoms in the molecule or ion.

If we consider NF3, we can easily see that nitrogen is less electronegative than fluorine, hence nitrogen is the central atom in the molecule. We can also observe from the molecular model that three atoms of fluorine were attached to the central atom. Hence there are three atoms attached to the central atom in the molecule NF3.

Water contains 2 polar bonds and the molecule is polar
True
False

Answers

TRUEEEEEEEEEEEEEEEEEEEEEEE

A weather system moving through the American Midwest produced rain with an average pH of 5.02. By the time the system reached New England, the rain it produced had an average pH of 4.66. How much more acidic was the rain falling in New England?

Answers

Answer:

The rain falling in New England is 2.29 times more acidic than the one in the American Midwest.

Explanation:

The acidity of a solution depends on the concentration of H⁺ ions ([H⁺]). We can calculate this concentration from the pH using the following expression.

pH = -log ([H⁺])

American Midwest

pH = -log ([H⁺])

5.02 = -log ([H⁺])

[H⁺] = antilog (-5.02) = 9.55 × 10⁻⁶ M

New England

pH = -log ([H⁺])

4.66 = -log ([H⁺])

[H⁺] = antilog (-4.66) = 2.19 × 10⁻⁵ M

The ratio of concentrations is:

$(2.19 * 10^(-5) M )/(9.55 * 10^(-6) M) =2.29$

The rain falling in New England is 2.29 times more acidic than the one in the American Midwest.

The change in the acidity of the rain falling in New England is mathematically given as

r=2.29

What is the change in the acidity of the rain falling in New England?

Question Parameter(s):

American midwest pH of 5.02

New England pH of 4.66

Generally, the equation for the pH is mathematically given as

pH = -log ([H⁺])

Therefore American midwest

pH = -log ([H⁺])

5.02 = -log ([H⁺])

[H⁺] = antilog (-5.02) = 9.55 × 10⁻⁶ M

and New england

pH = -log ([H⁺])

4.66 = -log ([H⁺])

[H⁺] = antilog (-4.66) = 2.19 × 10⁻⁵ M

In conclusion, the concentration ratio is

$r=(2.19 * 10^(-5) M )/(9.55 * 10^(-6) M)$

r=2.29

Answers

Answer:

The law is observed in the given equation.

Explanation:

CaCO₃ + 2HCI → CaCI₂ +H₂O + CO₂

In order to find out if the law of conservative mass is followed, we need to count how many atoms of each element are there in both sides of the equation:

Ca ⇒ 1 on the left, 1 on the right.

C ⇒ 1 on the left, 1 on the right.

O ⇒ 3 on the left, 3 on the right.

H ⇒ 2 on the left, 2 on the right.

Cl ⇒ 2 on the left, 2 on the right.

As the numbers for all elements involved are the same, the law is observed in the given equation.

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