list any five items that can be found in a Science portfolio​

Answer:

c2-c

Explanation:

A mental that lost two electrons just joking I don't k ow but I think it ain't the first o e because of comment sense to the 5th power

Answer:

A and B

Explanation:

metal form ion by lossing elecron

non metal form ion by gaining electron

Answer:

[tex]m_{Au_{31}}=6106.07 g[/tex]

Explanation:

Hello,

In this case, since one gold atom has a mass of 196.97 g, for the given gold cluster that has 31 gold atoms, the mass will be:

[tex]m_{Au_{31}}=31*196.97g\\\\m_{Au_{31}}=6106.07 g[/tex]

Whereas we consider the mass of all the 31 gold atoms to compute the overall mass.

Best regards.

Answer:

Explanation:

Atomic mass of Au = 197u

So, mass of 1 Au31 cluster (in u)

[tex]= 197 * 31\\\\= 6107u[/tex]

So, mass of 1 Au31 cluster (in gram)

[tex]= 6107 * 1.66*10^{-24}\\\\= 1.014*10^{-20}grams[/tex]

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Answer:

the wavelength of radiation emitted  is [tex]\mathbf{\lambda= 2169.62 \ nm}[/tex]

Explanation:

The energy of the Bohr's hydrogen atom can be expressed with the formula:

[tex]\mathtt{E_n =- \dfrac{13.6\ ev}{n^2}}[/tex]

For n = 7:

[tex]\mathtt{E_7 =- \dfrac{13.6\ ev}{7^2}}[/tex]

[tex]\mathtt{E_7 =-0.27755 \ eV}[/tex]

For n = 4

[tex]\mathtt{E_4=- \dfrac{13.6\ ev}{4^2}}[/tex]

[tex]\mathtt{E_4 =- 0.85\ eV}[/tex]

The  electron goes from the n = 7 to the n = 4, then :

[tex]\mathtt{E_7-E_4 = (-0.27755 - (-0.85) ) \ eV}[/tex]

[tex]\mathtt{= 0.57245\ eV}[/tex]

Wavelength of the radiation emitted:

[tex]\mathtt{\lambda= \dfrac{hc}{0.57245 \ eV}}[/tex]

where;

hc  = 1242 eV.nm

[tex]\mathtt{\lambda= \dfrac{1242 \ eV.nm }{0.57245 \ eV}}[/tex]

[tex]\mathbf{\lambda= 2169.62 \ nm}[/tex]

Answer:

1. the study of the structure, properties, and interactions of materials and the changes they undergo

2. the study of matter and energy and how they interact

3. the systematic study and analysis of facts

4. a process to test a proposed solution to a problem

Click again to see term

Until the 1600s, people who investigated problems related to our understanding of the natural world were called _____.

Click ca

Explanation:

Net ionic equation is calculated by the eliminating spectator ion. The net ionic products are potassium ion and bromine ion.

Net ionic equation:

It shows the ionic products of the products that has been precipitated by the dissociation of reactant in aquas solution.

Aqueous solutions of potassium hydroxide and ammonium bromide combined to form a precipitate KBr.

The dissociation reaction of  

[tex]\bold {KBr(aq) \rightarrow K^+(aq) + Br^-(aq)}[/tex]

Aquas potassium hydroxide dissociates into potassium ion and bromine ion.

Therefore, the net ionic products are potassium ion and bromine ion.

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Answer:

the transfer of heat energy within the atmosphere, hydrosphere, and the earth's surface and interior occurs as a result of radiation, convection, and conduction. ocean currents play a significant role in transferring this heat toward the poles.

Answer:

The amount of dissolved gases in the body. Have a good day! =)

Explanation:

Answer:

the answer will be B.1 and 3 only

Explanation:1. O2- means it has in total 10 electrons as atomic no of oxygen is 8

again F- has also 10 electrons as flourine has atomic no 9

and neon  also  has 10 electrons

incase of 3rd no S2- has 18 electrons as atomic no of sulfur is 16 as its an ion so it will take 2 more electrons

argon is an inert gas so it has 18 electrons and potassium has 19 electrons eliminating 1 we will 18

# incase of the second option lithium + contains 2 electrons and Na+ contains 10 and K+ contains 18 electrons so all of them are not same

ANS is 1 and 3 only(B.i and iii) only

thank u

Answer:

to prevent the chemical from reacting with ions present in water

Explanation:

water contains dissolved ions such as calcium from minerals and soluble rocks

Answer:

1) If life exists in other planets

2) What type of living organisms exists on other planets

3) Predominant atoms/mixture/compound that exists on other planets

Explanation:

Earth and space science is the study of the earth, the space and other planets within the universe.

The following questions can be answered by studying earth and space science

1) If life exists in other planets

Over the years, there has been several reports suggesting this but a deep inquiry into this with scientific data can assist in putting the matter to rest.

2) What type of living organisms exists on other planets

If there is/are lives on other planets, what type of life/living organisms exist there. Are they plants? Are they animals? Or do we have a new classification for them?

3) Predominant atom/mixture/compound that exists on other planets

What is the most popular substance in each of these planets? The role of this substance in the survival of the planet or it's inhabitants also needs to be studied

Answer:

d. 0.121 M HC2H3O2 and 0.116 M NaC2H3O2

Explanation:

Hello,

In this case, since the pH variation is analyzed via the Henderson-Hasselbach equation:

[tex]pH=pKa+log(\frac{[Base]}{[Acid]} )[/tex]

We can infer that the nearer to 1 the ratio of of the concentration of the base to the concentration of the acid the better the buffering capacity. In such a way, since the sodium acetate is acting as the base and the acetic acid as the acid, we have:

a. [tex]\frac{[Base]}{[Acid]}=\frac{0.497M}{0.365M}=1.36[/tex]

b. [tex]\frac{[Base]}{[Acid]}=\frac{0.217M}{0.521M}=0.417[/tex]

c. [tex]\frac{[Base]}{[Acid]}=\frac{0.713M}{0.821M}=0.868[/tex]

d. [tex]\frac{[Base]}{[Acid]}=\frac{0.116M}{0.121M}=0.959[/tex]

Therefore, the d. solution has the best buffering capacity.

Regards.

Answer:

Groups

Explanation:

1 A, 3 B, and 7 A are examples of  group number on the periodic table

Group 1A are  the alkali metals includes lithium sodium and potassium.

Group 3B in most periodic tables, lanthanum and actinium are considered to be a part of Group3B.

Group 7A are the halogens includes chlorine ,bromine and iodine.

They are all soft, silver metals. Due to their low ionization energy, these metals have low melting points and are highly reactive. The group 1A elements with their ns1 valence electron configurations are very active metals. They lose their valence electrons very readily.

All the Group 3B elements are rather soft, silvery-white metals, although their hardness increases with atomic number. They have higher ionization energies than the Group 1A and 2A elements, and are ionized to form a 3+ charges.

They have very high electronegativity. They have seven valence electrons . They are highly reactive, especially with alkali metals and alkaline earths metals.

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Answer:

7.89 g

Explanation:

Step 1: Write the balanced equation

S₈ + 16 F₂(g) → 8 SF₄

Step 2: Calculate the moles corresponding to 2.34 g of S₈

The molar mass of S₈ is 256.52 g/mol.

[tex]2.34g \times \frac{1mol}{256.52g} = 9.12 \times 10^{-3} mol[/tex]

Step 3: Calculate the moles of SF₄ produced from 9.12 × 10⁻³ mol of S₈

The molar ratio of S₈ to SF₄ is 1:8. The moles of SF₄ produced are 8/1 × 9.12 × 10⁻³ mol = 0.0730 mol

Step 4: Calculate the mass corresponding to 0.0730 moles of SF₄

The molar mass of SF₄ is 108.07 g/mol.

[tex]0.0730 mol \times \frac{108.07 g}{mol} = 7.89 g[/tex]

Answer:

1.34x10⁻⁵ mol / L is the solubility of AgCl

Explanation:

Ksp of AgCl is defined as:

AgCl(s) ⇄ Ag⁺(aq) + Cl⁻(aq)

If 0.010M of AgCl is added, some Ag⁺ and Cl⁻ will be produced until:

Ksp = [Ag⁺] [Cl⁻]

Ksp for AgCl = 1.8x10⁻¹⁰ (Taken from ALEKS Data tab):

Some Ag⁺ and Cl⁻ are produced, you can take this "some" as X:

[Ag⁺] = X

[Cl⁻] = X

Where X is the amount of AgCl that dissolvesin water. X = solubility:

Ksp = 1.8x10⁻¹⁰ = [Ag⁺] [Cl⁻]

1.8x10⁻¹⁰ = [X] [X]

1.8x10⁻¹⁰ = X²

X =

Answer:

The correct answer is : 0.025.

Explanation:

The precipitation reaction is as follows in this procedure:

NiSO4.6H20 (aq) + Na2CO3.10H2O (aq)⇒ NiCO3 (s) + Na2SO4 (aq) + 16H2O

So, the number of moles of reactants can be found by

number of moles : mass used/ molar mass

n (NiSO4×6H2O) = (6.57 g)/(262.84 g/mol) = 0.025 mol (for Ni)

n (Na2CO3×10H2O) = (7.15 g)/(286.14 g/mol) = 0.0250 mol (for CO3).

Thus, The maximum number of moles of NiCO3 that can form is 0.025 mol.

Answer:

1.66 moles.

Explanation:

We'll begin by calculating the number of mole in 23.2 g of nitrogen gas, N2.

This is illustrated below:

Molar mass of N2 = 2x14 = 28 g/mol

Mass of N2 = 23.2 g

Mole of N2 =.?

Mole = mass /Molar mass

Mole of N2 = 23.2/28

Mole of N2 = 0.83 mole

Next, we shall determine the number of mole in 23.2 g of Hydrogen gas, H2.

This is illustrated below:

Molar mass of H2 = 2x1 = 2 g/mol

Mass of H2 = 23.2 g

Mole of H2 =?

Mole = mass /Molar mass

Mole of H2 = 23.2/2

Mole of H2 = 11.6 moles

Next, the balanced equation for the reaction. This is given below:

N2 + 3H2 —> 2NH3

From the balanced equation above,

1 mole of N2 reacted with 3 moles of H2 to produce 2 moles of NH3.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

1 mole of N2 reacted with 3 moles of H2.

Therefore, 0.83 moles will react with = (0.83 x 3) = 2.49 moles of H2.

From the calculations made above, we can see that only 2.49 moles out of 11.6 moles of H2 is required to react completely with 0.83 mole of N2.

Therefore, N2 is the limiting reactant.

Finally, we shall determine the maximum amount of NH3 produced from the reaction.

In this case, we shall use the limiting reactant because it will give the maximum yield of NH3 since all of it is consumed in the reaction.

The limiting reactant is N2 and the maximum amount of NH3 produced can be obtained as follow:

From the balanced equation above,

1 mole of N2 reacted to produce 2 moles of NH3.

Therefore, 0.83 mole of N2 will react to produce = (0.83 x 2) = 1.66 moles of NH3.

Therefore, the maximum amount of NH3 produced from the reaction is 1.66 moles.

Answer:

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Explanation:

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Answer:

0.64 M

Explanation:

Step 1: Given data

Step 2: Calculate the final concentration of the dextrose solution

We want to prepare a diluted solution from a concentrated one. We can find the final concentration using the dilution rule.

C₁ × V₁ = C₂ × V₂

C₂ = C₁ × V₁ / V₂

C₂ = 1.0 M × 16 mL / 25 mL

C₂ = 0.64 M

Answer:

B, adding an indicator to a solution to test its acidity

Explanation:

Answer:

B - Adding an indicator to a solution to test its acidity

Explanation:

Took the test

Answer:

Pressure of hydrogen gas = 695.2 mmHg

Explanation:

Given:

Water temperature = 22°C

Pressure inside the tube = 715 mmHg

Find:

Pressure of hydrogen gas

Computation:

Using vapor pressure of water table

Water pressure at 22°C = 19.8 mmHg

Pressure inside the tube = Pressure of hydrogen gas + Water pressure at 22°C

715 = Pressure of hydrogen gas + 19.8

Pressure of hydrogen gas = 715 - 19.8

Pressure of hydrogen gas = 695.2 mmHg

Answer:

70

Explanation:

100/20 =5

5 x each day over 2 weeks (14 days) = 70

The vials are necessary for the entire treatment is 70 vials

We have to first calculate the total number of antibiotics taken by the patient in two weeks, then find the total vials neccessary for the entire treatment.

The total antibiotics for a period of two weeks = (100 mg/day)(2 weeks)

Total = 100 mg/day×14 days

Total = 1400 mg for two weeks.

If the antibiotics contain vials in 20 mg/vials of the drugs,

Then the vials necessary for the entire treatment is

vial for the entire treatment  = 1400/20

total vials = 70 vials

Hence the vials are necessary for the entire treatment is 70 vials

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Answer:

D) in chemical changes and physical changes.

Explanation:

Mass and energy are conserved in chemical changes and physical changes. Therefore, option D is correct.

Physical modifications are those that affect a chemical substance's form but not its chemical content. Physical changes may normally be used to separate compounds into chemical elements or simpler compounds, but they cannot be used to separate mixtures into their component compounds.

A change in physical attributes accompanies a physical change. Melting, turning into a gas, changing strength or durability, changing crystal structure or texture, and changing shape, size, color, volume, or density are a few examples of physical qualities.

In a physical change, the substance's shape or appearance changes, but the type of matter it contains stays the same. In contrast, when matter undergoes a chemical transformation, at least one new substance with novel features is created.

Thus, option D is correct.

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Explanation:

Moles of 0.042 L of Pure water = 0.042 / 22.4

[tex]Moles \: of \: 0.042 L \: of \: Pure \: water \: = 1.87 \times {10}^{ - 3} [/tex]

[tex]Molecules \: in \: 0.042 L \: of \: Pure \: water \: = moles \times avogadro \: number[/tex]

Molecules in 0.042 L of Pure Water

= 1.126 * 10^21

Answer:

diffraction

Explanation:

Louis de Broglie in 1924 proposed the idea of wave particle duality. In his proposition, matter could exhibit wavelike properties.

Electrons are generally regarded as particles. Electrons may also display wavelike properties such as diffraction patterns. Generally, diffraction is regarded as a property of waves.

Hence, electron diffraction effect owes to the wavelike nature of of electrons when they are passed near matter.

Sir Joseph John Thomson OM PRS (18 December 1856 – 30 August 1940) was a British physicist and Nobel Laureate in Physics, credited with the discovery of the electron, the first subatomic particle to be discovered.

The volume of KOH required for the neutralization of acid has been 38.9 ml. Thus, option B is correct.

Neutralization reaction has resulted in the formation of salt and water with the reaction of acid and base.

In the neutralization reaction, the strength of the acid and base can be given by:

Molarity of acid [tex]\times[/tex] Volume of acid = Molarity of base [tex]\times[/tex] Volume of base

Given, the molarity of KOH base = 0.124 M

The volume of acid (HCl) = 23.4 ml

Molarity of acid (HCl) = 0.206 M.

Substituting the values:

0.206 [tex]\times[/tex] 23.4 = 0.124 × Volume of base (KOH)

Volume of base (KOH) = 38.874 ml.

Volume of KOH = 38.9 ml

The volume of KOH required for the neutralization of acid has been 38.9 ml. Thus, option B is correct.

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Answer:

a. minimum voltage that must be supplied for a redox reaction to occur

c. always equal to Eanode - Ecathode

Explanation:

In an electrolytic cell;  The electromotive force(the maximum standard potential difference) of the cell formed by the system is defined as the standard electrode potential of the right handed electrode minus the standard electrode potential of the left hand electrode. (i.e [tex]\mathbf{E^{\theta}_{cell}=E_{anode} - E_{cathode}}[/tex] )

As we all known that the process by which chemical energy is being converted to electrical energy is called the Electrochemical cell. It consists of two half cells , an oxidation half cell reaction and a reduction half-cell reaction.The overall redox reaction results in a flow of electrons in an electric current which is produced by a minimum voltage.

Therefore, option a and c are both correct.