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mixas84 [53]
2 years ago
11

What information is shown in an atom’s electron dot diagram? the number of shells an atom has the number of electrons in each sh

ell of an atom the number of valence electrons an atom has the number of valence shells of an atom
Chemistry
2 answers:
tatiyna2 years ago
4 0
The number of valence electrons an atom has  
Aleks [24]2 years ago
3 0

The answer on Edge is,

C. the number of valence electrons an atom has

Please vote <u>brainliest</u> (:

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Glyceraldehyde-3-phosphate is an important intermediate molecule in the cell's metabolic pathways because __________.
olga55 [171]

Answer:

Present in both catabolic and anabolic pathways

Explanation:

Glyceraldehyde-3-phosphate abbreviated as G3P occurs as intermediate in glycolysis and gluconeogenesis.

In photosynthesis, it is produced by the light independent reaction and acts as carrier for returning ADP, phosphate ions Pi, and NADP+  to the light independent pathway. Photosynthesis is a anbolic pathway.

In glycolysis, Glyceraldehyde-3-phosphate is produced by breakdown of fructose-1,6 -bisphosphate. Further Glyceraldehyde-3-phosphate converted to pyruvate and pyruvate is further used in citric acid cycle for energy production. Therefore, it is used in catabolic pathway too.

Glyceraldehyde-3-phosphate is an important intermediate molecule in the cell's metabolic pathways because it is  present in both catabolic and anabolic pathways.

4 0
2 years ago
When 5.00 g of Al2S3 and 2.50 g of H2O are reacted according to the following reaction: Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H
Debora [2.8K]

Answer:

Y=58.15\%

Explanation:

Hello,

For the given chemical reaction:

Al_2S_3(s) + 6 H_2O(l) \rightarrow 2 Al(OH)_3(s) + 3 H_2S(g)

We first must identify the limiting reactant by computing the reacting moles of Al2S3:

n_{Al_2S_3}=5.00gAl_2S_3*\frac{1molAl_2S_3}{150.158 gAl_2S_3} =0.0333molAl_2S_3

Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:

n_{Al_2S_3}^{consumed}=2.50gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molAl_2S_3}{6molH_2O}=0.0231mol  Al_2S_3

Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:

m_{Al(OH)_3}=0.0231molAl_2S_3*\frac{2molAl(OH)_3}{1molAl_2S_3}*\frac{78gAl(OH)_3}{1molAl(OH)_3} =3.61gAl(OH)_3

Finally, we compute the percent yield with the obtained 2.10 g:

Y=\frac{2.10g}{3.61g} *100\%\\\\Y=58.15\%

Best regards.

7 0
2 years ago
Can someone do this for me i wasent there when we learned it
Anna11 [10]
I’ll do the first two for you.
4 0
3 years ago
4. Substances that existed before mixing-still exist-after substances-are-mixed----
pantera1 [17]
False; Because when 2 substances are mixed another substance is created.
5 0
2 years ago
What is the maximum amount of Kl that can be dissolved in 100 g H2O at 20C
olganol [36]

Answer: azada calva

Explanation:

7 0
3 years ago
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