For the oxidation–reduction reaction equation

2Sc+3Br2⟶2ScBr3

indicate how many electrons are transferred in the formation of one formula unit of product.

Answer:

LETTER

1. D. PURE SUBSTANCES CAN BE FURTHER BROKEN DOWN INTO SIMPLER SUBSTANCES.

2.A

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Answer:

The fourth atom.

Explanation:

Number of valence electrons in each of the four atoms:

Without referring to a periodic table, the first, second, and fourth atoms would all belong to the alkali-earth metal elements. The third atom would belong to a noble gas element. (The four choices are beryllium, magnesium, argon, and calcium, respectively.)

Noble gas elements are highly stable. Thus, the third atom (argon) would be the least reactive among all four choices.

On the other hand, alkali-earth metals are reactive since they tend to lose their valence electrons. The reactivity of an alkali-earth atom depends on its ionization energy (the energy required to take a valence electron away from the atom.)

Since these atoms are in the same group, ionization energy would be smaller for atoms with a larger radius. In a given group, atoms with more filled electron shells would have a larger radius.

The first atom has two filled main shells, the second has three, whereas the fourth has four.

Thus, among the three alkali-earth metals in the choices:

[tex]\begin{aligned}&\text{Number of filled electron shells:} \\ & a < b < d\end{aligned}[/tex].

[tex]\begin{aligned}&\text{Atomic radius:} \\ & a < b < d\end{aligned}[/tex].

[tex]\begin{aligned}&\text{Ionization energy:} \\ & a > b > d\end{aligned}[/tex].

[tex]\begin{aligned}&\text{Reactivity:} \\ & a < b < d\end{aligned}[/tex].

The third choice (a noble gas) is the least reactive among the four choices:

[tex]\begin{aligned}&\text{Reactivity:} \\ & c < a < b < d\end{aligned}[/tex].

Overall, the fourth choice would be the most reactive among the four choices.

This  problem provides the molar mass and radius of a metal that has an FCC unit cell and the density is required.

Firstly, we begin with the formula that relates the aforementioned variables and also includes the Avogadro's number and the volume of the unit cell:

[tex]\rho=\frac{Z*M}{V*N_A}[/tex]

Whereas Z stands for the number of atoms in the unit cell, M the molar mass, V the volume and NA the Avogadro's number. Next, since FFC is able to hold 4 atoms and M and NA are given. Next, we calculate the volume of the atom in the unit given the radius in meters:

[tex]V=a^3=(2*1.92x10^-10m*\sqrt{2} )^3=1.60x10^{-28}m^3/atom[/tex]

And finally the required density in g/cm³:

[tex]\rho=\frac{4*241.5g/mol}{1.60x10^{-28}m^3/atom*6.022x10^{23}atom/mol} \\\\\rho=10025739g/m^3=10.03 g/cm^3[/tex]

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Answer:

ooooooooh .,.,.,.,.,.,

Answer:

True

TruePhysical change – A change in the size, shape, color, or state of matter of a substance. No new substance is produced.

When the value of secondary quantum number (l) = 1, the magnetic quantum number can be from -1 to +1. The answer would be false.

The angular quantum number (l) can take any integer between 0 and the value of the principal quantum number (n) less by 1.

In other words, l can be from 0 to n-1.

The magnetic quantum number (m), on the other hand, can take any value between -l to +l.

For example, if l = 1, m can be -1,0, and +1.

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Answer:

Explanation:

Answer:

78n

Explanation:

The output force exerted on the nail in the claw is equal to 78 N which has a mechanical advantage of 5.2.

The mechanical advantage can be demonstrated as the ratio of the output force to the Input force. The mechanical advantage of any machine can be expressed in the form of the ratio of the forces utilized to do the work.

The ratio of the resistance to the effort is said to be the actual mechanical advantage which will be less. The efficiency of a machine can be evaluated by equating the ratio of the output to its input.

Given, the input force = 15 N

The mechanical advantage of the hammer = 5.2

Mechanical advantage = Output force/ Input force

5.2 = Output/15

Output force = 15 ×5.2 = 78 N

Therefore, the force is exerted on the nail in the claw is equal to 78 N.

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Answer:

10

Explanation:

Equation: d = m/v

100/10 = 10

Answer:A versatile Ideal Gas Laws calculator with which you can calculate the pressure, volume, quantity (moles) or temperature of an ideal gas, given the other three. Free online gas law calculator a.k.a. PV = nRT calculator which accepts different input metric units such as temperature in celsius, fahrenheit, kelvin; pressure in pascals, bars, atmospheres; volume in both metric and imperial units ...

Explanation:

[tex]\tiny\implies Molarity = \dfrac{no. \: of \: moles \: of \: solute \times 1000}{volume \: of \: the \: solution \: (in \: ml)} [/tex]

[tex]\tiny\implies 0.50 = \dfrac{no. \: of \: moles \: of \: solute \times 1000}{15}[/tex]

[tex]\tiny\implies no. \: of \: moles \: of \: solute \times 1000 = 0.50 \times 15[/tex]

[tex]\tiny\implies no. \: of \: moles \: of \: solute \times 1000 = 7.5[/tex]

[tex]\tiny\implies no. \: of \: moles \: of \: solute = \dfrac{7.5}{1000} [/tex]

[tex]\tiny\implies \bf no. \: of \: moles \: of \: solute = 0.075 \: mol[/tex]

Answer:

where is the question????????????

Answer:

B

Explanation:

as temperature rises, the particles gain kinetic energy (they will move faster) so if temperatures stays constant, so will the movement or vibration of the particles

Answer:

A. because there both high pitched voice and low pitched ( girls have sweet, shrill ,high pitched good voice ) where as boys have horase ,low pitched medium voice )

Answer:

go get the stuff.

Explanation:

A sample of crude sodium iodide was analyzed by the following balanced reaction. The oxidation number of S in SO₄²⁻ is +6.

8 I⁻ + 6 H₂O + SO₄²⁻ → 4 I₂ + H₂S + 10 OH⁻

Let's consider the following unbalanced redox reaction.

I⁻ + SO₄²⁻ → I₂ + H₂S

The corresponding half-reactions are:

I⁻ → I₂

SO₄²⁻ → H₂S

We will perform the mass balance adding OH⁻ and H₂O where appropriate.

2 I⁻ → I₂

6 H₂O + SO₄²⁻ → H₂S + 10 OH⁻

Then, we will perform the charge balance adding electrons where appropriate.

2 I⁻ → I₂ + 2 e⁻

8 e⁻ + 6 H₂O + SO₄²⁻ → H₂S + 10 OH⁻

Finally, we will multiply the first half-reaction by 4 and the second by 1, and add them.

4 × (2 I⁻ → I₂ + 2 e⁻)

1 × (8 e⁻ + 6 H₂O + SO₄²⁻ → H₂S + 10 OH⁻)

------------------------------------------------------------

8 I⁻ + 6 H₂O + SO₄²⁻ → 4 I₂ + H₂S + 10 OH⁻

A sample of crude sodium iodide was analyzed by the following balanced reaction. The oxidation number of S in SO₄²⁻ is +6.

8 I⁻ + 6 H₂O + SO₄²⁻ → 4 I₂ + H₂S + 10 OH⁻

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Answer:

A, B, C, D

Explanation:

The water cycle is included in the atmosphere (water vapor in the air), the geosphere (water collection), the hydrosphere (water on earth: lakes, oceans, rivers), and the biosphere (where water interacts with all living things).

In an ecosystem, the water cycle is included in all the spheres that is biosphere,geosphere,hydrosphere and atmosphere.

Ecosystem is defined as a system which consists of all living organisms and the physical components with which the living beings interact. The abiotic and biotic components are linked to each other through nutrient cycles and flow of energy.

Energy enters the system through the process of photosynthesis .Animals play an important role in transfer of energy as they feed on each other.As a result of this transfer of matter and energy takes place through the system .Living organisms also influence the quantity of biomass present.By decomposition of dead plants and animals by microbes nutrients are released back in to the soil.

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In this case, the problem is asking for the balance of a redox reaction in acidic media, in which nickel is reduced to a metallic way and nitrogen oxidized to an ionic way.

Thus, according to the given information, it turns out possible for us to balance this equation in acidic solution by firstly setting up the half reactions:

[tex]Ni^{2+}+2e^-\rightarrow Ni^0\\\\N^{3-}H_4^++3H_2O\rightarrow N^{5+}O_3^-+8e^-+10H^+[/tex]

Next, we cross multiply each half-reaction by the other's carried electrons:

[tex]8Ni^{2+}+16e^-\rightarrow 8Ni^0\\\\2N^{3-}H_4^++6H_2O\rightarrow 2N^{5+}O_3^-+16e^-+20H^+[/tex]

Finally, we add them together to obtain:

[tex]8Ni^{2+}+2N^{3-}H_4^++6H_2O\rightarrow 8Ni^0+2N^{5+}O_3^-+20H^+[/tex]

Which can be all simplified by a factor of 2 to obtain:

[tex]4Ni^{2+}+N^{3-}H_4^++3H_2O\rightarrow 4Ni^0+N^{5+}O_3^-+10H^+\\\\4Ni^{2+}(aq)+NH_4^+(aq)+3H_2O(l)\rightarrow 4Ni(s)+NO_3^-(aq)+10H^+(aq)[/tex]

Hence, the coefficients in front of Ni and H⁺ are 4 and 10 respectively.

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