I need the answer to this multiple-choice

Answer: D) Matter becomes more dense as it moves from gas to solid state

A gas expands to fill up the container of whatever shape or size. As it cools to a solid, the molecules get closer together and don't move as much (the molecules still will vibrate and move somewhat even in solid form). So effectively the volume of the object shrinks while the amount of mass stays the same. This is why the density increases when going from gas to solid.

Answer:

Endothermic

Explanation:

In order to melt the ice cube, heat is required, so the process is endothermic. Endothermic reaction In an endothermic reaction, the products are higher in energy than the reactants. Therefore, the change in enthalpy is positive, and heat is absorbed from the surroundings by the reaction.

There are total three laws of newtons, first law of newtons, second law of newton and third law of newton. Therefore, when a golf club is subjected to an imbalanced force, its velocity changes, which is according to newton’s second law.

According to Newton's Second Law of Motion, acceleration (gaining speed) occurs when a force acts on a mass (object). Riding a bicycle is a great example of this rule of motion in action. The mass is represented by your bicycle. The force is generated by your leg muscles pulling on the pedals of your bicycle.

The ball remains balanced on the tee due to the balanced forces of gravity downhill and upward. Force equals mass multiplied by acceleration (f=ma). When a golf club is subjected to an imbalanced force, its velocity changes.

Therefore, when a golf club is subjected to an imbalanced force, its velocity changes, which is according to newton’s second law.

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Answer:

your answer should be B.273 K

Explanation:

Answer:

Heat can travel from one place to another in three ways: Conduction, Convection and Radiation. Both conduction and convection require matter to transfer heat. If there is a temperature difference between two systems heat will always find a way to transfer from the higher to lower system.

Explanation:

okay so the  Answer is c

Answer:

2

Explanation:

BIonary bi means 2

i hope this helps :)

Answer:

The answer is true

Explanation: Because The law of multiple proportions states that when two elements react to form more than one compound, a fixed mass of one element will react with masses of the other element in a ratio of small, whole numbers.

Answer:

BeO will have the greatest mass composition of oxygen

Explanation:

To solve this problem, we would find the mass composition of oxygen in each of the given sample.

To find this mass composition, let us derive the molecular mass of the given species;

  K₂O₂ = 2(39) + 2(16) = 110g/mol

  Na₂O = 2(23) + 16 = 62g/mol

   BeO = 9 + 16  = 25g/mol

Now; we know that mass of each sample is 1g;

mass composition of O in K₂O₂ = [tex]\frac{32}{110}[/tex] x  1= 0.29g

mass composition of O in Na₂O = [tex]\frac{16}{62}[/tex] x 1 = 0.26g

mass composition of O in BeO = [tex]\frac{16}{25}[/tex] x 1 = 0.64g

We can see that BeO will have the greatest mass composition of oxygen.

The mass composition of oxygen has been highest in BeO. Thus, option 3 is correct.

The molecular mass has been the mass of each element that has been present in the compound. The molecular mass of given compounds have been:

= 2 (39) + 2 (16)

= 110 g/mol

= 2 (23) + 16

= 62 g/mol

= 9 + 16

= 23 g/mol

The mass composition of oxygen in 1 gram of the following compounds have been:

Mass composition = [tex]\rm \dfrac{Mass\;of\;Oxygen}{Molecular\;mass\;of\;compound}\;\times\;weight\;of\;sample[/tex]

[tex]\rm K_2O_2[/tex] = 0.29 grams

[tex]\rm Na_2O[/tex] = 0.26 grams

BeO = 0.64 grams.

The mass composition of oxygen has been highest in BeO. Thus, option 3 is correct.

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Answer:

6 atoms

Explanation:

Answer:

Explanation:

2NaNO₃ = 2NaNO₂ + O₂

NaNO₃    =     NaNO₂   +    1/2 O₂

- 468 kJ           -369 kJ           0  kJ

enthalpy of decomposition reaction

= - 369 - ( - 468 ) kJ

= 99 kJ / mol .      

Answer:

Neutral atoms can be turned into positively charged ions by removing one or more electrons.

Explanation:

If there is an atom that has 9 protons and 9 electrons, removing an electron from the atom will gain a postive charge.

Answer:

Q = 7.7 KJ

Explanation:

The complete question carries the following data:

Specific Heat Capacity of Liquid State = C(l) = 1.45 J/mol.°C

Specific Heat Capacity of Gaseous State = C(g) = 0.65 J/mol.°C

Boiling Temperature = Tb = 88.5°C

Heat of Vaporization = ΔH(vap) = 1.23 KJ/mol

Now, first we calculate the heat required (Q₁) to raise he temperature to boiling point:

Q₁ = n*C(l)*ΔT₁

where,

n = no. of moles = 5.31 mol

ΔT₁ = Temperature difference from 40°C to Tb = 88.5°C - 40°C = 48.5°C

Therefore,

Q₁ = (5.31 mol)(1.45 J/mol.°C)(48.5°C)

Q₁ = 373.4 J = 0.37 KJ

Now, we calculate the heat required (Q₂) to change its phase:

Q₂ = nΔH(vap)

Q₂ = (5.31 mol)(1.23 KJ/mol)

Q₂ = 6.53 KJ

Now, we calculate the heat required (Q₃) to raise he temperature to boiling point:

Q₃ = n*C(g)*ΔT₃

where,

n = no. of moles = 5.31 mol

ΔT₃ = Temperature difference from Tb to 113°C = 113°C - 88.5°C = 24.5°C

Therefore,

Q₃ = (5.31 mol)(0.65 J/mol.°C)(24.5°C)

Q₃ = 84.5 J = 0.08 KJ

So, the total heat required to convert 5.31 mol of this pure substance from a liquid at 40°C to a gas at 113°C is:

Q = Q₁ + Q₂ + Q₃

Q = 0.37 KJ + 6.53 KJ + 0.08 KJ

Q = 7.7 KJ

The total heat required to convert the pure substance from liquid to gas at the given temperature is 814 J.

The given parameters;

The total heat required to convert the pure substance from liquid to gas at the given temperature is calculated as follows;

[tex]Q = nc_l \Delta \theta + nc_g \Delta \theta \\\\ \Delta \theta = (113 - 40) = 73^0 C \\\\Q = (5.31 \ mol \times 1.45 \frac{J}{mol .\ ^0C} \times 73\ ^0C) + (5.31 \ mol \times 0.65 \frac{J}{mol .\ ^0C} \times 73 \ ^0C)\\\\Q = 814 \ J[/tex]

Thus, the total heat required to convert the pure substance from liquid to gas at the given temperature is 814 J.

"Your question is not complete, it seems to be missing the following information;"

How much heat would be required to convert 5.31 mol of a pure substance from a liquid at 40.0°C to a gas at 113.0°C? (molar heat capacity of liquid, C(l) is 1.45 J/mol.⁰C and molar heat capacity of gas, C(g) is 0.65 J/mol.⁰C).

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Answer:

Explanation:

M1V1=M2v2

M1=0.150M

M2=0.150M

V2=120*10^-3

V1=M2V2/M1

V1=0.150*120*10^-3/0.150

V1=0.12

0.12 mL volume of 0.150M 2M Li₂S solution is required to completely react with 120 mL of 0.150M Co(NO₃)₂ MCo(NO₃)₂.

With M1 and M2 representing the molarity of the solutions, expressed as mol/L or M, and V1 and V2 representing their respective volumes, this calculator may be used to calculate a missing number for the dilution equation.

A solution's concentration is determined by multiplying its volume by its molarity (M1V1 = M2V2) On both sides of the equation, the units should stay the same.

Here

M₁V₁=M₂V₂

M₁=0.150M

M₂=0.150M

V₂=120 × 10⁻³

V₁=M₂V₂/M₁

V₁=0.150 × 120 × 10⁻³/0.150

V₁=0.12.

Therefore, 0.12 mL volume of 0.150M 2M Li₂S solution is required to completely react with 120 mL of 0.150M Co(NO₃)₂ MCo(NO₃)₂.

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Answer:

X represents London dispersion forces, and Y represents dipole-dipole forces.

Explanation:

Answer:

The answer is Diffusion.

Answer:

Hypertonic

Explanation:

Answer:

Environmental cleanup includes removal of heavy metals and other toxic contaminants from the environment and the process is called remediation.

Answer: Therefore, the volume of a 0.155 M potassium hydroxide solution  is 56.0 ml

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

According to the neutralization law,

[tex]n_1M_1V_1=n_2M_2V_2[/tex]

where,

[tex]M_1[/tex] = molarity of [tex]HBr[/tex] solution = 0.338 M

[tex]V_1[/tex] = volume of [tex]HBr[/tex] solution = 25.7 ml

[tex]M_2[/tex] = molarity of [tex]KOH[/tex] solution = 0.155 M

[tex]V_2[/tex] = volume of [tex]KOH[/tex] solution = ?

[tex]n_1[/tex] = valency of [tex]HBr[/tex] = 1

[tex]n_2[/tex] = valency of [tex]KOH[/tex] = 1

[tex]1\times 0.338\times 25.7=1\times 0.155\times V_2[/tex]

[tex]V_2=56.0ml[/tex]

Therefore, the volume of a 0.155 M potassium hydroxide solution  is 56.0 ml

Answer:

ON

Explanation:

Frictional force is a force that opposes the motion of body. It is because of friction that motion can be controlled and even walking is made possible.

Now, on a frictionless surface, there is no opposition to motion. Therefore, to keep the body moving there is no force required.

Answer:

Mechanical

Explanation:

chemical

Explanation:

I took test 2020

Answer:

atomic number of 12 is MG

Explanation:

Magnesium

Answer:

1. a

Explanation:

2. c

Answer:

1 is A number two is C

Explanation:

i dont understand dont you already have the equation?

The net ionic equation for the acid-base reaction PH[tex]_3[/tex](aq) + HCl(aq) [tex]\rightarrow[/tex] PH[tex]_4[/tex]Cl(aq) is PH[tex]_3[/tex](aq) + H⁺ (aq) [tex]\rightarrow[/tex] PH[tex]_4[/tex]⁺ .

Chemical reaction equations can be written in a variety of ways. Unbalanced chemical equations, which identify the species involved, balanced chemical equations, which identify the number and type of species, molecular equations, that also express compounds as molecules rather than component ions, and net ionic equations, which only address the species which contribute to a reaction, are some of the most popular.

A chemical equation for just a process known as the net ionic equation only includes the species that are really involved in the reaction.

PH[tex]_3[/tex](aq) + HCl(aq) [tex]\rightarrow[/tex] PH[tex]_4[/tex]Cl(aq)

PH[tex]_3[/tex](aq) + H⁺ (aq)  + Cl⁻ (aq) [tex]\rightarrow[/tex] PH[tex]_4[/tex]⁺ (aq)  + Cl⁻ (aq)

PH[tex]_3[/tex](aq) + H⁺ (aq) [tex]\rightarrow[/tex] PH[tex]_4[/tex]⁺

Therefore,  the net ionic equation for the acid-base reaction is PH[tex]_3[/tex](aq) + H⁺ (aq) [tex]\rightarrow[/tex] PH[tex]_4[/tex]⁺ .

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a. The unit cell is the smallest group of atom which have overall symmetry of a crystal, and from which is the entire letters can be buled built up by repetition in 3 dimensions.

b. The volume(v) of the unit cell is equal to the cell edge length (a)cubed.

c. density of polonium is 9.32g/cm3.

Answer:

See explanation

Explanation:

The reaction is shown in the image attached. This reaction occurs by a carbocation (ionic) mechanism.

The lone pair on the nitrogen atom attacks the carbocation leading to the formation of the product as shown in the image attached.

Since the reaction occurs in a 1:1 ratio, two moles of reactants yields two moles of product.

The phosphate group of one nucleotide bonds covalently with the sugar molecule of the next nucleotide, and so on, forming a long polymer of nucleotide monomers. The sugar–phosphate groups line up in a “backbone” for each single strand of DNA, and the nucleotide bases stick out from this backbone. The carbon atoms of the five-carbon sugar are numbered clockwise from the oxygen as 1′, 2′, 3′, 4′, and 5′ (1′ is read as “one prime”). The phosphate group is attached to the 5′ carbon of one nucleotide and the 3′ carbon of the next nucleotide. In its natural state, each DNA molecule is actually composed of two single strands held together along their length with hydrogen bonds between the bases.

Answer:

Explanation:

The discovery that DNA is the prime genetic molecule, carrying all the hereditary information within chromosomes, immediately focused attention on its structure. It was hoped that knowledge

of the structure would reveal how DNA carries the genetic messages that are replicated when chromosomes divide to produce two identical copies of themselves. During the late 1940s and early 1950s, several research groups in the United States and in Europe engaged in serious efforts—both cooperative and rival—to understand how the atoms of DNA are linked together by covalent bonds and how the resulting molecules are arranged in three-dimensional space. Not surprisingly, there initially were fears that DNA might have very complicated and perhaps bizarre structures that differed radically from one gene to another. Great relief, if not general elation, was thus expressed when the fundamental DNA structure was found to be the double helix. It told us that all genes have roughly the same three-dimensional form and that the differences between two genes reside in the order and number of their four nucleotide building blocks along the complementary strands.

Now, some 50 years after the discovery of the double helix, this simple description of the genetic material remains true and has not had to be ap- preciably altered to accommodate new findings. Nevertheless, we have come to realize that the structure of DNA is not quite as uniform as was first thought. For example, the chromosome of some small viruses have single-stranded, not double-stranded, molecules. Moreover, the precise orientation of the base pairs varies slightly from base pair to base pair in a manner that is influenced by the local DNA sequence. Some DNA se- quences even permit the double helix to twist in the left-handed sense, as opposed to the right-handed sense originally formulated for DNA’s general structure. And while some DNA molecules are linear, others are circular. Still additional complexity comes from the supercoiling (further twisting) of the double helix, often around cores of DNA-binding proteins.

Likewise, we now realize that RNA, which at first glance appears to be very similar to DNA, has its own distinctive structural features. It is principally found as a single-stranded molecule. Yet by means of intra-strand base pairing, RNA exhibits extensive double-helical character and is capable of folding into a wealth of diverse tertiary structures. These structures are full of surprises, such as non-classical base pairs, base-backbone interactions, and knot-like configurations. Most remarkable of all, and of profound evolutionary significance, some RNA molecules are enzymes that carry out reactions that are at the core of information transfer from nucleic acid to protein.

Clearly, the structures of DNA and RNA are richer and more intricate than was at first appreciated. Indeed, there is no one generic structure for DNA and RNA. As we shall see in this chapter, there are in fact vari- ations on common themes of structure that arise from the unique physi- cal, chemical, and topological properties of the polynucleotide chain