All categories

2 years ago

What reason best explains why metals are shiny

2 answers:

8 0

Metals are shiny because of their electrons. These electrons produce light when they come in contact with light. I hope this helps.

Elanso [62]2 years ago

4 0

The shiny appearance of a metal is most closely related to the metal's highly mobile valence electrons. In metallic bonds, the mobile electrons surrounding the positive ions are called a(n) electron sea. As light strikes the surface of a metal, the electrons in the electron sea absorb and re-emit the light.
I hope this helps.

You might be interested in

Calculate the density of carbon dioxide at STP

Artist 52 [7]

Answer:

Density = mass/volume

= 44/22.4

= 1.96 gram/liter

The density of the Carbon Dioxide at S.T.P. (Standard Temperature and Volume) is 1.96 gram/liter.

5 0

2 years ago

10. How many g of Cu(OH)2 can be made from 9.1 x 1025 atoms of O?

il63 [147K]

Molar mass Cu(OH)₂ = 97.561 g/mol

97.561 g Cu(OH)₂ --------------- 6.02x10²³ atoms
? g Cu(OH)₂ -------------------- 9.1x10²⁵ atoms

mass = 9.1x10²⁵ * 97.561 / 6.02x10²³

mass = 8.87x10²⁷ / 6.02x10²³

mass = 14734.2 g

hope this helps!

8 0

3 years ago

What material is the source for commercial production of each of the following elements:(a) aluminum;

AleksandrR [38]

The material which is used as source for commercial production aluminum is bauxite.

The aluminum can be extracted from bauxite ore by the process of Bayer process.

In the Bayer process, bauxite ore is heated in the pressure vessel along with a caustic soda solution (sodium hydroxide) at a temperature between 150 to 200 °C. At this temperatures, the aluminium is dissolved in the solution as sodium aluminate in the extraction process. After separation of the residue by filtering, when the liquid is cooled gibbsite is precipitated and then it is seeded with fine-grained aluminum hydroxide crystals from previous extractions. The precipitation take 7-19 days without the addition of seed crystals.

This extraction process converts the aluminium oxide to soluble sodium aluminate, NaAlO2, which afterward converted into aluminum hydroxide and then into aluminum oxide.

Thus, we concluded that the material which is used as source for commercial production aluminum is bauxite ore.

learn more about ore:

brainly.com/question/10306443

#SPJ4

3 0

1 year ago

The concentration of protein in a urine sample is calculated to be 2.77 μg/mL. What is the concentration of this solution in uni

maxonik [38]

Answer:

The concentration of this solution in units of pounds per gallon is $2.776*10^{-5} \frac{lb}{gal}$

Explanation:

Units of measurement are established models for measuring different quantities. The conversion of units is the transformation of a quantity, expressed in a certain unit of measure, into an equivalent one, which may or may not be of the same system of units.

In this case, the conversion of units is carried out knowing that 1 μg are equal to 2.205*10⁻⁹ Lb and 1 mL equals 0.00022 Gallons. So

$2.77 \frac{ug}{mL} = \frac{2.77 ug}{mL}$

If 1 μg equals 2.205*10⁻⁹ lb, 2.77 μg how many lb equals?

$lb=\frac{2.77ug*2.205*10^{-9}lb }{1ug}$

lb=6.10785*10⁻⁹

So, 2.77 μg= 6.10785*10⁻⁹ lb

Then:

$2.77 \frac{ug}{mL} = \frac{2.77 ug}{mL}=\frac{6.10785*10^{-9}lb }{mL} =\frac{6.10785*10^{-9}lb }{0.00022 gal} =\frac{6.10785*10^{-9}lb }{0.00022 gal}$

You get:

$2.77 \frac{ug}{mL} = 2.776*10^{-5} \frac{lb}{gal}$

The concentration of this solution in units of pounds per gallon is $2.776*10^{-5} \frac{lb}{gal}$

6 0

2 years ago

What volume of 0.307 m naoh must be added to 200.0ml of 0.425m acetic acid (ka = 1.75 x 10-5 ) to produce a buffer of ph = 4.250

Blababa [14]

The buffer solution target has a pH value smaller than that of pKw (i.e., pH < 7.) The solution is therefore acidic. It contains significantly more protons $\text{H}^{+}$ than hydroxide ions $\text{OH}^{-}$ . The equilibrium equation shall thus contain protons rather than a combination of water and hydroxide ions as the reacting species.

Assuming that $x \; \text{L}$ of the 0.307 $\text{mol} \cdot \text{dm}^{-3}$ sodium hydroxide solution was added to the acetic acid. Based on previous reasoning, $x$ is sufficiently small that acetic acid was in excess, and no hydroxide ion has yet been produced in the solution. The solution would thus contain $0.2000 \times 0.425 - 0.307 \; x = 0.085 - 0.307 \; x$ moles of acetic acid and $0.307 \; x$ moles of acetate ions.

Let $\text{HAc}$ denotes an acetic acid molecule and $\text{Ac}^{-}$ denotes an acetate ion. The RICE table below resembles the hydrolysis equilibrium going on within the buffer solution.

$\begin{array}{lccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\end{array}$

The buffer shall have a pH of 4.250, meaning that it shall have an equilibrium proton concentration of $10^{4.250}\; \text{mol}\cdot \text{dm}^{-3}$ . There were no proton in the buffer solution before the hydrolysis of acetic acid. Therefore the table shall have an increase of $10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3}$ in proton concentration in the third row. Atoms conserve. Thus the concentration increase of protons by $10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3}$ would correspond to a decrease in acetic acid concentration and an increase in acetate ion concentration by the same amount. That is:

$\begin{array}{lcccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\text{C} & - 10^{-4.250} & & +10^{-4.250} & & +10^{-4.250} \\\text{E} & 0.085 - 10^{-4.250} - 0.307 \; x& & 10^{-4.250} & & 10^{-4.250} + 0.307 \; x\end{array}$

By definition:

$\text{K}_{a} = [\text{H}^{+}] \cdot [\text{Ac}^{-}] / [\text{HAc}]\\\phantom{\text{K}_{a}} = 10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x)$

The question states that

$\text{K}_{a} = 1.75 \times 10^{-5}$

such that

$10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x) = 1.75 \times 10^{-5}\\6.16 \times 10^{-5} \; x = 1.48 \times 10^{-6}\\x = 0.0241$

Thus it takes $0.0241 \; \text{L}$ of sodium hydroxide to produce this buffer solution.

6 0

2 years ago