The list shows the results of six athletes competing in the long jump, which has a target distance of 20 feet.

On the basis of chemical structure, which solid will most easily conduct electricity?

Marizza181 [45]

Answer:

Fe

Explanation:

The electrical conductivity depends mainly on the type of chemical bonds between the atoms of a compound.

In the case of MgF2, FeCl3 and FeO3, these have the type of ionic bond. This means that in the atoms of the compound there is an electron transfer, to keep eight electrons in the outermost layer and thus resemble the electronic configuration of the inert gas closest to each of the two elements, due to this ions of opposite charges are formed that are held together by electrostatic forces. These types of compounds are good conductors of electricity, however, to have this property, they must be dissolved in water or molten.

In the case of Fe, however, the type of union between atoms is metallic. In this type of junction, valence electrons are quite free inside the metal, which makes it easy for them to move. For this reason, this compound will conduct electricity in a solid state.

6 0

2 years ago

A 3.5 kg iron shovel is left outside through the winter. The shovel, now orange with rust, is rediscovered in the spring. Its ma

Anton [14]

Answer:

a

Explanation:

8 0

1 year ago

A student uses visible spectrophotometry to determine the concentration of CoCl2(aq) in a sample solution. First the student pre

k0ka [10]

Answer:

$4\times 10^{-9} J$ is the approximate energy of one photon of this light.

Explanation:

Energy of the photon can be calculated by

$E=h\nu=\frac{h\times c}{\lambda}$ (Planck's equation)

where,

E = energy of photon

h = Planck's constant = $6.63\times 10^{-34}Js$

c = speed of light = $3\times 10^8m/s$

$\lambda$ = wavelength of light =

$\nu$ = frequency of the light

we have , $\lambda =510 nm=510\times 10^{-9}m$

Now put all the given values in the above formula, we get the energy of the photons.

$E=\frac{(6.63\times 10^{-34}Js)\times (3\times 10^8m/s)}{510\times 10^{-9}m}$

$E=3.9\times 10^{-19}J\approx 4\times 10^{-9} J$

$4\times 10^{-9} J$ is the approximate energy of one photon of this light.

5 0

2 years ago

Ron and Hermione begin with 1.50 g of the hydrate copper(II)sulfate ∙ x-hydrate (CuSO4 ∙ xH2O), where x is an integer. Part of t

Gwar [14]

Answer

5

Explanation:

We can go about this using the percentage compositions.

First, we calculate the percentage composition of the copper sulphate. This is obtainable by using the mass.

0.96/1.5 * 100 = 64%

Hence the percentage by mass of the water present is 36%

The molar mass of the anhydrous sulphate is 64 + 32 +4(16) = 160g/mol

The molar mass of the water is 2(1) + 16 = 18g/mol

Not forgetting that it is in multiples of x, the total molar mass of the water is 18x moles

The total mass of the copper sulphate hydrate is 160+ 18x

Now how do we get x? Like it is said earlier, the percentage composition is constant.

Hence, 64/100 * (160 + 18x) = 160

16000 = 64(160 + 18x)

16000 = 10,240 + 1152x

16,000 - 10,240 = 1152x

1152x = 5760

x = 5760/1152

x = 5

7 0

2 years ago

What is the density of 2.5 g of gaseous sulfur held at 130. kPa and 10.0 degrees Celsius?

worty [1.4K]

Use ideal gas equation: pV = nRT

Now pass n to mass: n = mass / MM .... [MM is the molar mass]

pV = [mass/MM]*RT =>mass/V = [p*MM] / RT and mass / V = density

p= 130 kPa = 130,000 Pa = 130,00 joule / m^3
T = 10.0 ° + 273.15 = 283.15 k
MM of sulfur (S) = 32 g/mol = 32000 kg/mol

density = 130,000 Pa * 32000kg/mol / [8.31 joule / mol*k * 283.15 k] = 1.77*10^6 kg/m^3 = 1.77 g/L ≈ 1.8 g/L

Then, I do not get any of the option choices.

Is it possbile that the pressure is 13.0 kPa instead 130. kPa? If so the answer would be 18 g/L

Note that the mass is not used. You do not need it unless you are asked for the volume, which is not the case.

4 0

2 years ago

Read 2 more answers