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Otrada [13]
3 years ago
9

Anybody wanna help me with number 22? (Chemistry)

Chemistry
1 answer:
Inga [223]3 years ago
8 0

Answer:

Rutherford atomic model of atom

Explanation:

lord lutherford

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Which salt shows the least change in solubility from 0 ⁰ – 100 ⁰ C?
MrRissso [65]

Answer:

From the solubility curve, which salt shows the least change in solubility as the temperature increases? NaCl

Explanation:

5 0
2 years 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
Amino acids are the building blocks of proteins. The simplest amino acid is glycine (H2NCH2COOH). Draw a Lewis structure for gly
VashaNatasha [74]

Answer:

Explanation:

The lewis structure (indicating all the atoms and patterns provided as hint in the question) of glycine can be seen in the attachment below. While the chemical structure of glycine can be seen below

         H

          |

H₂N - C - C =O

          |      \

         H      OH

The structure (of glycine) above provides a "fair idea" of how the lewis structure will be.

3 0
3 years ago
He standard reduction potentials of lithium metal and chlorine gas are as follows: reaction reduction potential (v) li+(aq)+e−→l
AURORKA [14]

 2 Li(s)  +Cl₂→  2 Li⁺ (aq)  + 2Cl⁻ (aq)

The cell potential   of  the reaction above   is +4.40V

<em><u>calculation</u></em>

Cell  potential  =∈° red - ∈° oxidation

in  reaction above  Li  is  oxidized from oxidation state 0 to +1 therefore the∈° oxid = -3.04

  Cl  is  reduce   from oxidation  state 0 to -1 therefore  the ∈°red = +1.36 V

cell  potential is therefore = +1.36 v -- 3.04  = + 4.40 V

3 0
3 years ago
How many grams of sodium sulfide can be produced when 45.3 g Na react with 105 g S?
RoseWind [281]
The chemical reaction would be as follows:

<span>2Na + S → Na2S

We are given the amount of the reactants to be used in the reaction. We use these to calculate the amount of product. We do as follows:

45.3 g Na ( 1 mol / 22.99 g ) = 1.97 mol Na
105 g S ( 1 mol / 32.06 g ) = 3.28 mol S

The limiting reactant would be Na. We calculate as follows:

1.97 mol Na ( 1 mol Na2S / 2 mol Na ) (78.04 g / mol ) = 76.87 g Na2S produced</span>
8 0
3 years ago
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