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3 years ago

What will happen if ultraviolet light shines on metal

Chemistry

2 answers:

Romashka [77]3 years ago

8 0

Answer:

if this doesnt help then im sorry...

Explanation:

When light is shined on certain metals, electrons may be knocked off the metal. This is called the photoelectric effect. By assuming that light is quantized, Einstein was able to explain the photoelectric effect. ... When UV light of a long wavelength and low frequency is shined on the zinc nothing happens.

Nezavi [6.7K]3 years ago

3 0

Ah yay! Something I can answer.

Answer:

When UV light is shined upon certain metals something called the photoelectric effect occurs. The photoelectic effect is when electrons are discharged from the metal.

Explanation:

Metals contain many electrons and when the electrons are given enough energy they can leave the metal. The electrons that escape the metal are called photoelectrons; hence why this effect is called the photoelectic effect.

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Place the following molecular ions in order from smallest to largest bond order: <img src="https://tex.z-dn.net/?f=C_2%5E%7B2%2B

podryga [215]

Answer:

The correct answer is option 1 "F(2-) "less than" C2(2+) "less than" O(2-) "less than" N(2-)".

Explanation:

Bond order is a term used in chemistry to identify the number of electrons involved in making the bonding between two or more atoms in a molecule. Bond order is equal to half the difference between the number of bonds in the atoms minus the number of antibonds. Therefore, the correct order of bond order from smallest to largest is as follows:

1. F(2-): equals to 0.5 (it has 8 bonds and 7 antibonds)

2. C2(2+): equals to 1 (it has 4 bonds and 2 antibonds)

3. O(2-): equals to 2 (it has 8 bonds and 4 antibonds)

4. N(2-): equals to 2.5 (it has 8 bonds and 3 antibonds)

5 0

3 years ago

If 1.50 L of 0.780 mol/L sodium sulfide is mixed with 1.00 L of a 3.31 mol/L lead(II) nitrate solution, what mass of precipitate

NARA [144]

Answer:

336.1 g of PbS precipitate

Explanation:

The equation of the reaction is given as;

Na2S(aq) + Pb(NO3)2(aq) ----> 2NaNO3(aq) + PbS(s)

Ionically;

Pb^2+(aq) + S^2-(aq) -----> PbS(s)

Number of moles of sodium sulphide= concentration of sodium sulphide × volume of sodium sulphide

Number of moles of sodium sulphide= 0.780 × 1.5 = 1.17 moles

Number of moles of lead II nitrate= concentration of lead II nitrate × volume of lead II nitrate

Number of moles of lead II nitrate= 3.31× 1.00= 3.31 moles

Then we determine the limiting reactant. The limiting reactant yields the least amount of product.

Since 1 moles of sodium sulphide yields 1 mole of lead II sulphide

1.17 moles of sodium sulphide also yields 1.17 moles of lead II sulphide

Hence sodium sulphide is the limiting reactant.

Thus mass of precipitate formed= amount of lead II sulphide × molar mass of sodium sulphide

Molar mass of lead II sulphide= 287.26 g/mol

Mass of lead II sulphide = 1.17 moles × 287.26 g/mol

Mass of lead II sulphide= 336.1 g of PbS precipitate

4 0

3 years ago

Question 6 Jackson visited the Grand Canyon with his family. He noticed that when he stood at the edge of the canyon and yelled

Novosadov [1.4K]

Answer:

Explanation:

7 0

3 years ago

Data Collection

vladimir2022 [97]

Answer:

For a): The mass percent of naphthalene in the original sample is 31.00 %.

For b): The mass percent of 3-nitroaniline in the original sample is 32.03 %.

For c): The mass percent of benzoic acid in the original sample is 29.97 %.

For d): The total percent recovered is 93.00 %.

Explanation:

Percentage by mass is defined as the ratio of the mass of a substance to the mass of the solution multiplied by 100. The formula used for this is:

$\text{Percent by mass}=\frac{\text{Mass of substance}}{\text{Mass of a solution}} \times 100$ ......(1)

a):

Mass of naphthalene = 0.483 g

Mass of the sample = 1.558 g

Plugging values in equation 1:

$\%\text{ mass of naphthalene}=\frac{0.483 g}{1.558}\times 100\\\\\%\text{ mass of naphthalene}=31.00 \%$

b):

Mass of 3-nitroaniline = 0.499 g

Mass of the sample = 1.558 g

Plugging values in equation 1:

$\%\text{ mass of 3-nitroaniline}=\frac{0.499 g}{1.558}\times 100\\\\\%\text{ mass of 3-nitroaniline}=32.03 \%$

c):

Mass of benzoic acid = 0.467 g

Mass of the sample = 1.558 g

Plugging values in equation 1:

$\%\text{ mass of benzoic acid}=\frac{0.467 g}{1.558}\times 100\\\\\%\text{ mass of benozic acid}=29.97 \%$

d):

Total mass recovered = [0.483 + 0.499 + 0.467] = 1.449 g

Mass of the sample = 1.558 g

Plugging values in equation 1:

$\text{Total percent recovered}=\frac{1.449 g}{1.558}\times 100\\\\\text{Total percent recovered}=93.00\%$

8 0

3 years ago

Object E has a mass of 4,800 kilograms. Object F has a mass of 600 kilograms. The weight of Object F will be ________ times the

vovangra [49]

Answer:The weight of Object F will be $\frac{1}{8}$ times the weight of Object E if both objects are placed on the same planet.

Explanation:

Mass of the object E, $M_E$ = 4800 kg

Mass of the object F, $M_F$ == 600 kg

Since they are on the same planet same gravitation force will be exerted on them by the gravity of the planet.

$\frac{M_E}{M_F}=\frac{4800 kg}{600 kg}=\frac{8}{1}$

$M_F=M_E\times \frac{1}{8}$

The weight of Object F will be $\frac{1}{8}$ times the weight of Object E if both objects are placed on the same planet.

8 0

3 years ago