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3 years ago

Please answer with an actual answer and not just put a random word ^-^

Chemistry

1 answer:

WARRIOR [948]3 years ago

6 0

Which eclipse was modeled when the large ball was between the small ball and the light?

The model is a "Lunar Eclipse" (If it was talking about the earth, then yes, it is a lunar eclipse).

Which eclipse was modeled when the small ball was between the large ball and the light?

The model is a "Solar Eclipse".

What does the large ball represent?

The earth.

What does the small ball represent?

The moon.

What does the light source represent?

The sun.

Hope this helps!~ <3

(I can't draw so sorry.)

You might be interested in

How many planets are in our galaxy? how many are inhabitable by humans?

viktelen [127]

...

100 billion planets are in our galaxy.

About 60 billion planets are inhabitable by humans.

Hope this helps!

3 0

3 years ago

Secondary bile salts lack a hydroxyl (OH) group when compared to primary bile salts. What is the consequence of the loss of this

Slav-nsk [51]

Answer:

When secondary bile salts (or acids) lack a OH, they are reabsorbed

Explanation:

The lost of a hydroxyl group (OH) in the bile salts structure means less hydro solubility. Thus, they are not eliminated, conversely, the are reabsorbed for intestine and they go back to the liver.

I hope my answer helps

7 0

3 years ago

Lithium nitride reacts with water to produce ammonia and lithium hydroxide according to the equation, Li3N(s) + 3H2O (L) -->

Gemiola [76]

Answer:

0.480 grams

Explanation:

Li₃N(s) + 3D₂O (L) --------------------------> ND₃(g) + 3LiOD (aq)

1 : 3 : 1 : 3

Number of moles (n) = Mass in gram/ Molar Mass

Mass of ND₃ = 160 mg

= 0.16 g

Molar mass of ND₃= [14 + (3 x 2.014 )]

= 14 + 6.042

= 20.042 g/mol

Number of moles of ND₃ = 0.16/20.042

= 0.007983 moles

From the reaction equation, the mole ratio between Heavy water (D₂O ) and ND₃ is 3: 1.

This implies that the number of moles of Heavy water (D₂O ) required

= 3 x 0.007983 moles

= 0.023949 moles

Molar mass of Heavy water (D₂O )= [(2.014 x 2) + 16]

= 20.028 g/mol

Mass in grams of Heavy water (D₂O )= Number of moles x Molar mass

= 0.023949 x 20.028

= 0.4797 grams

≈ 0.480 grams

3 0

3 years ago

Calculate the energy (in kj/mol) required to remove the electron in the ground state for each of the following one-electron spec

Bess [88]

Explanation:

$E_n=-13.6\times \frac{Z^2}{n^2}ev$

where,

$E_n$ = energy of $n^{th}$ orbit

n = number of orbit

Z = atomic number

a) Energy change due to transition from n = 1 to n = ∞ ,hydrogen atom .

Z = 1

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{1^2}{1^2}eV=-13.6 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{1^2}{(\infty)^2}eV=0$

Let energy change be E for 1 atom.

$E=E_{\infty}-E_1=0-(-13.6 eV)=13.6 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 13.6 eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 13.6 \times 1.60218\times 10^{-22} kJ/mol$

$E'=1,312.17 kJ/mol$

The energy required to remove the electron in the ground state is 1,312.17 kJ/mol.

b) Energy change due to transition from n = 1 to n = ∞ , $B^{4+}$ atom .

Z = 5

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{5^2}{1^2}eV=-340 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{5^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-340eV)=340 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 340eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 340\times 1.60218\times 10^{-22} kJ/mol$

$E'=32,804.31 kJ/mol$

The energy required to remove the electron in the ground state is 32,804.31 kJ/mol.

c) Energy change due to transition from n = 1 to n = ∞ , $Li^{2+}$ atom .

Z = 3

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{3^2}{1^2}eV=-122.4 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{3^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-122.4 eV)=122.4 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 122.4 eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 122.4\times 1.60218\times 10^{-22} kJ/mol$

$E'=11,809.55 kJ/mol$

The energy required to remove the electron in the ground state is 11,809.55 kJ/mol.

d) Energy change due to transition from n = 1 to n = ∞ , $Mn^{24+}$ atom .

Z = 25

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{25^2}{1^2}eV=-8,500 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{25^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-8,500 eV)=8,500 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 8,500eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 8,500 \times 1.60218\times 10^{-22} kJ/mol$

$E'=820,107.88 kJ/mol$

The energy required to remove the electron in the ground state is 820,107.88 kJ/mol.

4 0

3 years ago

Can cells that are haploid (single set of chromosomes ) undergo meiosis

Dima020 [189]

Answer:

Yes they can

Haploid cells are mostly sex cells and these cells usually undergo meiosis.

Hope this helps.

5 0

3 years ago