Which characteristics of metal atoms help explain why valence electrons in a metal are delocalized?

Answers

Answer 1

Answer: In a metal, the electrons are considered free because there are more electrons than there should be and also transition metals are willing to accept and give up electrons from their d-orbitals. The d-orbital shell are in a relatively high energy state which means they are loosely bound to the atom where they can freely move around.

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A rock can be broken down into different kinds of substances by physical processes. No chemical reactions are needed to separate different parts of a rock into pure substances. This is because a rock is a(n) .
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Which element has the greatest density at stp? 1. carbon 2. copper 3. chlorine 4. calcium

Answers

Copper will have the greatest density, followed by calcium, another metal. Carbon will be the third most dense followed by chlorine, which, as a gas, will be MUCH less dense than the other 3

Final answer:

Among carbon, copper, chlorine, and calcium, copper has the highest density at standard temperature and pressure.

Explanation:

The element with the greatest density at standard temperature and pressure (STP) among the options given (carbon, copper, chlorine, calcium) is copper. Density, in elemental terms, is determined by the amount of mass packed in a given volume. Copper (Cu) has a density of approximately 8.96 g/cm^3 at STP, which is greater than that of carbon, chlorine, and calcium.

For comparison:
Carbon(C) has a significantly lower density at about 2.26 g/cm^3, chlorine (Cl) is a gas at STP and has a very low density of 0.003214 g/cm^3, and calcium (Ca) has a density of about 1.55 g/cm^3 at STP.

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A 115.0-g sample of oxygen was produced by heating 400.0 g of potassium chlorate.What is the percent yield of oxygen in this chemical reaction?

Answers

The balanced chemical reaction will be:

2KClO3 = 2KCl + 3O2

We are given the amount of potassium chlorate being burned. This will be our starting point.

400.0 g KClO3 1 mol KClO3/ 122.55 g KClO3) (3 mol O2/2 mol KClO3) ( 32.00 g O2/1mol O2) = 156.67 g O2

Percent yield = actual yield / theoretical yield x 100

Percent yield =115.0 g / 156.67 g x 100

Percent yield = 73.40 %

In which laboratory process could a student use 0.10 M NaOH(aq) to determine the concentration of an aqueous solution of HBr?(1) chromatography
(2) decomposition of the solute
(3) evaporation of the solvent
(4) titration

Answers

Answer:

4) titration

Explanation:

Titration is a standard process used in a laboratory to determine the concentration of an unknown analyte. A titrant of known concentration is gradually added to a known volume of the analyte in the presence of a suitable indicator. The end of the titration is marked by a color change of the analyte.

The given example is that of an acid(HBr) - base(NaOH) titration which can be represented by the following equation:

NaOH + HBr → NaBr + H2O

Thus 1 mole of acid gets neutralized by 1 mole of the base to form 1 mole of the salt (NaBr)

Let M1 and V1 are the molarity and volume of the base (NaOH). Here, the molarity of NaOH is known = M1 = 0.10 M and the volume, V1 corresponds to the end point in the titration.

M2 and V2 are the molarity and volume of HBr. Here, V2 is known whereas M2 needs to be determined.

Based on the reaction stoichiometry:

moles of NaOH = moles of HBr

$M1*V1=M2*V2\n\nTherefore,\n\nM2 = (M1*V1)/(V2)$

(4) titration, the process of using an aqueous solution of known concentration to determine the concentration of another solution of unknown concentration, is your answer.

Please answer me ASAPA technique that uses a porous barrier (separation with pores) to separate heterogeneous mixtures is _______.

a. distillation

b. chromatography

c. filtration

d. crystallization

Answers

Answer:

c.filtration is the answer

Which one of these is NOT a specialized area of Earth science?A. astronomy B. environmental science C. technology D. oceanograph

Answers

the answer is c technology

71. The density of osmium (the densest metal) is 22.57 g/cm². If a1.00-kg rectangular block of osmium has two dimensions of
4.00 cm x 4.00 cm, calculate the third dimension of the block.

Answers

Answer:

y = 2.77 cm

Answer: The third dimension is 2.77 cm

Final answer:

The question deals with density in physics, specifically regarding the densest metal, osmium. The dimensions of an osmium block are partially given, and the task is to find the third dimension. However, without the mass of the block, it is impossible to give a complete solution.

Explanation:

The subject of the question involves understanding the concept of density, which is the mass of an object divided by its volume. The density of osmium is given as 22.57 g/cm³. The student is also given the dimensions of the osmium block, which forms a rectangular prism with known length and width (4.00 cm x 4.00 cm), but unknown height. D = m/v or v = m/D can be used to find the third dimension of the block (height).

However, to use this formula to get a complete solution, the mass of the block must be known, which is not provided in the question. Without this crucial information, we can't solve this problem correctly.

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