All categories

4 years ago

PLEASEE ANSWERRR FASTTT

Chemistry

2 answers:

IceJOKER [234]4 years ago

8 0

Answer:

last one

Explanation:

The elements classified as metalloids are boron, silicon, germanium, arsenic, antimony, tellurium, and polonium.

Vsevolod [243]4 years ago

5 0

Answer:

Explanation:

has silicone and polonium the others dont

You might be interested in

Which is the best method for organizing information collected in an investigation? Plz help

Elden [556K]

Construct a table i think

7 0

4 years ago

Read 2 more answers

When metallic sodium is dissolved in liquid sodium chloride, electrons are released into the liquid. These dissolved electrons a

qaws [65]

Answer:

The edge of the length is $\mathbf{L = 8.54 \times 10^{-10} \ m}$

Explanation:

From the given information:

The associated energy for a particle in three - dimensional box can be expressed as:

$E_n = \dfrac{h^2}{8mL^2}(n_x^2+n_y^2+n_z^2)$

here;

h = planck's constant = $6.626 \times 10^{-34} \ Js$

$n_i$ = the quantum no in a specified direction

m = mass (of particle)

L = length of the box

At the ground state $n_x = n_y = n_z=1$

The energy at the ground state can be calculated by using the formula:

$E_1 =\dfrac{3h^2}{8mL^2}$

At first excited energy level, one of the quantum values will be 2 and the others will be 1.

Thus, the first excited energy will be: 2,1,1

∴

$E_2 =\dfrac{(2^2+1^2+1^2)h^2}{8mL^2}$

$E_2 =\dfrac{(4+1+1)h^2}{8mL^2}$

$E_2 =\dfrac{(6)h^2}{8mL^2}$

The transition energy needed to move from the ground to the excited state is:

$\Delta E= E_2 - E_1$

$\Delta E= \dfrac{6h^2}{8mL^2}- \dfrac{3h^2}{8mL^2}$

$\Delta E= \dfrac{3h^2}{8mL^2}}$ ----- (1)

Recall that:

the wavelength identified with the electronic transition is: 800 nm

800 nm = 8.0 × 10⁻⁷ m

However, the energy-related with the electronic transition is:

$\Delta E =\dfrac{hc}{\lambda}$

$\Delta E =\dfrac{6.626 \times 10^{-34} \times 2.99 \times 10^8}{8.0 \times 10^{-7} }$

$\Delta E =2.48 \times 10^{-19} \ J$

Replacing the value of $\Delta E$ in (1); then:

$2.48 \times 10^{-19}= \dfrac{3h^2}{8mL^2}}$

Making the edge length L the subject of the formula; we have:

$L = \sqrt{\dfrac{3h^2}{8m \times2.48 \times 10^{-19}} }$

$L = \sqrt{\dfrac{3\times (6.626 \times 10^{-34})^2}{8(9.1 \times 10^{-31} ) \times2.48 \times 10^{-19}} }$

$\mathbf{L = 8.54 \times 10^{-10} \ m}$

Thus, the edge of the length is $\mathbf{L = 8.54 \times 10^{-10} \ m}$

5 0

3 years ago

In the titration of 50.0 mL of HCl of unknown concentration, the phenolphthalein indicator present in the colorless solution tur

KIM [24]

Answer:

The molarity of HCl is 0.138 M

Explanation:

The titration reaction is as follows:

2HCl + Ca(OH)₂ → CaCl₂ + 2H₂O

When no more HCl is left, the small excess of Ca(OH)₂ added will cause the pH to rise and the indicator will turn. At this point, the number of moles of Ca(OH)₂ added will be the same as half the number of moles of HCl since 1 mol Ca(OH)₂ reacts with 2 moles HCl. Then:

At the endpoint:

moles Ca(OH)₂ = moles HCl / 2

Knowing the number of moles of Ca(OH)₂ added, we can calculate the number of moles of the acid:

mol Ca(OH)₂ = Volume added * concentration of Ca(OH)₂

mol Ca(OH)₂ = 0.0265 l * 0.130 mol/l = 3.45 x 10⁻³ mol Ca(OH)₂

The number of moles of HCl will be:

mol HCl = 2 * 3.45 x 10⁻³ mol = 6.89 x 10⁻³ mol HCl

This number of moles was present in 50.0 ml, then, in 1000 ml:

mol of HCl in 1000 ml = 6.89 x 10⁻³ mol HCl * (1000ml / 50ml) = 0.138 mol

Then:

Molarity HCl = 0.138 M

7 0

4 years ago

In column chromatography (microscale), after loading it with solvent and adsorbent and prior to loading the sample, what level s

Nikolay [14]

Answer:

Explanation:

You should allow the solvent to drop to the level of the adsorvent, so it would never run dry.

When you let your sample to run dry it will never finish to flow from the adsorbent depending of it polarity.

Water should not be used because it can dissolve the adsorbent.

You could use another technique to identify the compound, as an infrared or a ultraviolet detector. You can also, if you know the compounds, identify it for the retention time, for example, if you need to detect two compounds, one more polar than the other, and use a polar adsorbent and a non-polar solvent, the first compound to exit the column will be the less polar one, because it will have a bigger interaction with the solvent than the stationary phase (adsorbent) and will go faster, the second will be the more polar one, because it will have a bigger interaction with the stationary phase.

5 0

4 years ago

Predict where the largest jump between successive ionization energies occurs for Rb.

Irina18 [472]

Answer: confused

Explanation:

8 0

4 years ago