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2 years ago

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3. Scientific methods may include three steps of study as listed below. Explain each step in detail with a complete content rela

ted sentence. (ref: p.12-16) a. Hypothesis b. Theory c. Scientific law

1 answer:

victus00 [196]2 years ago

8 0

Answer:

Hypothesis is an assumption or idea about a particular topic or argument. An hypothesis should be one which is able to be tested and measurable to determine its authenticity.

A theory is an explanation of a scientific observation which has undergone series of experiments and is reproducible in any part of the world.

A law is simply a rule which gives an in depth explanation of a scientific finding. If new findings emerge the law could be changed or modified.

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Question 6 of 10

Levart [38]

Answer:

C. Solids made of molecules

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2 years ago

What does ClF3 stand for

Nostrana [21]

Answer:

CHLORINE TRIFLUORIDE

Explanation:

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2 years ago

An oxygen atom has a mass of 2.66 x 10^-23 g and a glass of water has a mass of 0.050kg. Use this information to answer the ques

fomenos

<h2>ANSWER OF EACH PART ARE GIVEN BELOW</h2>

Explanation:

A)

We know, each mole contains $N_A=$ $6.023 \times 10^{23}$ atoms.

It is given that mass of one oxygen atom is m= $2.66\times 10^{-23}\ g$ .

Therefore, mass of one mole of oxygen, $M=m\times N_A$ .

Putting value of n and $N_A$ ,

$M=2.66\times 10^{-23}\times 6.023\times 10^{23} \ gm\\M=16.0\ gm$

B)

Given,

Mass of water in glass=0.050 kg = 50 gm.

From above part mass of one mole of oxygen atoms = 16.0 gm.

Therefore, number of mole of oxygen equivalent to 50 gm oxygen $=\dfrac{50}{16}=3.1 \ moles.$

LEARN MORE :

Avogadro's number

brainly.com/question/12902286

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2 years ago

Compare which element would have larger first ionization energy: an alkali metal in Period 2 or an alkali metal in Period 4?

maria [59]

Answer:

An alkali metal present in period 2 have larger first ionization energy.

Explanation:

Ionization energy:

The amount of energy required to remove the electron from the atom is called ionization energy.

Trend along period:

As we move from left to right across the periodic table the number of valance electrons in an atom increase. The atomic size tend to decrease in same period of periodic table because the electrons are added with in the same shell. When the electron are added, at the same time protons are also added in the nucleus. The positive charge is going to increase and this charge is greater in effect than the charge of electrons. This effect lead to the greater nuclear attraction. The electrons are pull towards the nucleus and valance shell get closer to the nucleus. As a result of this greater nuclear attraction atomic radius decreases and ionization energy increases because it is very difficult to remove the electron from atom and more energy is required.

Trend along group:

As we move down the group atomic radii increased with increase of atomic number. The addition of electron in next level cause the atomic radii to increased. The hold of nucleus on valance shell become weaker because of shielding of electrons thus size of atom increased.

As the size of atom increases the ionization energy from top to bottom also decreases because it becomes easier to remove the electron because of less nuclear attraction and as more electrons are added the outer electrons becomes more shielded and away from nucleus. Thus alkali metal present in period 2 have larger ionization energy because of more nuclear attraction as compared to the alkali metal present in period 4.

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2 years ago

Identify and label the Brønsted-Lowry acid, its conjugate base, the Brønsted-Lowry base, and its conjugate acid in each of the f

julia-pushkina [17]

Explanation:

As per Brønsted-Lowry concept of acids and bases, chemical species which donate proton are called Brønsted-Lowry acids.

The chemical species which accept proton are called Brønsted-Lowry base.

(a) $HNO_3 + H_2O \rightarrow H_3O^+ + NO_3^-$

$HNO_3$ is Bronsted lowry acid and $NO_3^-$ is its conjugate base.

$H_2O$ is Bronsted lowry base and $H_3O^+$ is its conjugate acid.

(b)

$CN^- + H_2O \rightarrow HCN + OH^-$

$CN^-$ is Bronsted lowry base and HCN is its conjugate acid.

$H_2O$ is Bronsted lowry acid and $OH^-$ is its conjugate base.

(c)

$H_2SO_4 + Cl^- \rightarrow HCl + HSO_4^-$

$H_2SO_4$ is Bronsted lowry acid and $HSO_4^-$ is its conjugate base.

Cl^- is Bronsted lowry base and HCl is its conjugate acid.

(d)

$HSO_4^-+OH^- \rightarrow SO_4^{2-}+H_2O$

$HSO_4^-$ is Bronsted lowry acid and $SO_4^{2-}$ is its conjugate base.

OH^- is Bronsted lowry base and $H_2O$ is its conjugate acid.

(e)

$O_{2-}+H_2O \rightarrow 2OH^-$

$O_{2-}$ is Bronsted lowry base and OH- is its conjugate acid.

$H_2O$ is Bronsted lowry acid and OH- is its conjugate base.

6 0

2 years ago