Please help me with these?

put the contributions to the understanding of the atomic structure in order from most recent at the top to the earliest at the b

r-ruslan [8.4K]

Answer:

From Top to Bottom:

- Democritus coming up with the concept of an atom

- Dalton discovering that atoms are the smallest part of an element

- Rutherford discovering the nucleus of an atom

- Thomson discovering electrons

- Bohr modeling electrons orbiting the nucleus

- Schrodinger modeling electrons in the electron cloud

Explanation:

The best way to think about this is from the inside out. Democrats (who lived long before any of the other scientists mentioned) was the one who thought of the idea of the atom. - Therefore, this must be first because all other choices are elaborations on the idea that atoms exist. Next must be Dalton. Dalton saw atoms as "cannonballs" if you will; a solid mass. So then after that, Rutherford and his gold foil experiment (he discovered that some rays he shot through gold foil were deflected back; ie the existence of concentrated areas in an atom, ie the nucleus). Then we get into the information on electrons. We must start with discovery (Thomson). Heres where it gets complicated. Electrons don't actually orbit the nucleus, they exist in electron clouds. So it would be Bohr, who came up with the idea that electron exist outside the nucleus, then Schrodinger, who elaborated on Bohr's theory. Hope this helps!

Nat, Junior

Accel + AP Chem student

5 0

1 year ago

Lewis dot diagram for the Cs1+ ion

creativ13 [48]

Answer:

$Cs^+$

Explanation:

Cesium Lewis dot structure would look like this:

·Cs, because it only has one valence electron. But, since it has a plus, that means we lost an electron. So, we have to get rid of the dot and you have:

$Cs^+$

7 0

2 years ago

How many grams of aluminum are required to react with 35 mL of 2.0 M hydrochloric acid, HCl?__ HCl + __ Al --> __ AlCl3 + __

kiruha [24]

Answer:

0.6258 g

Explanation:

To determine the number grams of aluminum in the above reaction;

determine the number of moles of HCl
determine the mole ratio,
use the mole ratio to calculate the number of moles of aluminum.
use RFM of Aluminum to determine the grams required.

Moles of HCl

35 mL of 2.0 M HCl

2 moles of HCl is contained in 1000 mL

x moles of HCl is contained in 35 mL

$x \: mol \: = \: \frac{2 \: \times \: 35}{1000} \\ = 0.07 \: moles \:$

We have 0.07 moles of HCl.

Mole ratio

Balanced equation

6HCl(aq) + 2Al(s) --> 2AlCl3(aq) + 3H2(g)

Hence mole ratio = 6 : 2 (HCl : Al

but moles of HCl is 0.07, therefore the moles of Al;

$= \frac{2}{6} \times 0.07 \\ \: = 0.0233333 \: moles$

Therefore we have 0.0233333 moles of aluminum.

Grams of Aluminum

We use the formula;

$grams \: = moles \: \times \: rfm$

The RFM (Relative formula mass) of aluminum is 26.982g/mol.

Substitute values into the formula;

$= 0.0233333 \: moles \: \times \: 26.982 \: \frac{g}{mol} \\ = 0.625799 \: grams$

The number of grams of aluminum required to react with HCl is 0.6258 g.

3 0

2 years ago

If 65.5 moles of an ideal gas is at 9.15 atm at 50.30 °C, what is the volume of the gas?

qwelly [4]

To calculate for the volume, we need a relation to relate the number of moles (n), pressure (P), and temperature (T) with volume (V). For simplification, we assume the gas is an ideal gas. So, we use PV=nRT.

PV = nRT where R is the universal gas constant
V = nRT / P
V = 65.5 ( 0.08205 ) (273.15 + 50.30) / 9.15
V = 189.98 L

7 0

3 years ago

You observe 2 waves named alpha and beta. The alpha wave has a frequency of 5 Hz and the beta wave has a frequency of 2 Hz. Base

MAVERICK [17]

Answer:

shorter wavelength = alpha wave

Explanation:

Given that,

The alpha wave has a frequency of 5 Hz and the beta wave has a frequency of 2 Hz.

We need to compare the wavelengths of these two waves.

For alpha wave,

$\lambda_1=\dfrac{c}{f_1}\\\\\lambda_1=\dfrac{3\times 10^8}{5}\\\\=6\times 10^7\ m$

For beta wave,

$\lambda_2=\dfrac{c}{f_2}\\\\\lambda_2=\dfrac{3\times 10^8}{2}\\\\=15\times 10^7\ m$

From the above calculations, we find that the wavelength of the alpha wave is shorter than the wavelength of the beta wave.

8 0

3 years ago