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AfilCa [17]
3 years ago
11

The value of Δ G ° ' for the conversion of 3-phosphoglycerate to 2-phosphoglycerate (2PG) is + 4.40 kJ/mol . If the concentratio

n of 3-phosphoglycerate at equilibrium is 2.45 mM , what is the concentration of 2-phosphoglycerate? Assume a temperature of 25.0 ° C .
Chemistry
1 answer:
Mrac [35]3 years ago
8 0

Answer:The concentration of 2-phosphoglycerate is 0.415 mM

Explanation:

3-phosphoglycerate\rightleftharpoons 2-phosphoglycerate

Relation between standard Gibbs free energy and equilibrium constant follows:

\Delta G^o=-RT\ln K

where,

\Delta G^o = Standard Gibbs free energy = +4.40  kJ/mol = 4400 J/mol  (Conversion factor: 1kJ = 1000J)

R = Gas constant = 8.314J/K mol

T = temperature = 25^0C=(25+273)K=298 K

Putting values in above equation, we get:

4400J/mol=-(8.314J/Kmol)\times 298K\times \ln K

\ln K=-1.776

K=0.169

K=\frac{ 2-phosphoglycerate}{3-phosphoglycerate}

0.169=\frac{ 2-phosphoglycerate}{2.45mM}

2-phosphoglycerate}=0.415mM

Thus the concentration of 2-phosphoglycerate is 0.415 mM

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Which to which! Ahhhh.
jarptica [38.1K]

Answer:

1.a

2. b

Explanation:

Distillation is a process whereby a mixture of liquids having different vapor pressures is separated into its components. At first one might think that this would be quite simple: if you have a solution consisting of liquid A that boils at 50°C and liquid B with a boiling point of 90°C, all that would be necessary would be to heat the mixture to some temperature between these two values; this would boil off all the A (whose vapor could then be condensed back into pure liquid A), leaving pure liquid B in the pot. But that overlooks that fact that these liquids will have substantial vapor pressures at all temperatures, not only at their boiling points.

source: https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_Chem1_(Lower)/08%3A_Solutions/8.09%3A_Distillation

3 0
1 year ago
The reaction of methane with water to form carbon dioxide and hydrogen is nonspontaneous at 298 K. At what temperature will this
KIM [24]

Answer:

The system makes the transition from nonspontaneous to spontaneous at a temperature of 954.7 K.

Under 954.7 K the reaction is nonspontaneous; more than 954.7 K is the reaction spontaneous.

Explanation:

CH4(g) + 2H2O(g) ⇆ CO2(g) + 4H2(g)

CH4(g) H2O(g) CO2(g) H2(g) ΔH°f (kJ/mol): –74.87 –241.8 –393.5 0

ΔG°f (kJ/mol): –50.81 –228.6 –394.4 0

S°(J/K·mol): 186.1 188.8 213.7 130.7

ΔG<0 to be spontaneous

ΔG = ΔH- TΔS <0

ΔH = ∑nΔH(products) - ∑nΔH(reactant)

ΔH = (-393.5) - (–74.87 + 2*–241.8)

ΔH = 164.97 kJ = 164970 J

ΔS = ∑nΔS(products) - ∑nΔS(reactant)

ΔS = (213.7 + 4*130.7) - (186.1 + 2*188.8)

ΔS = 172.8 J

0 > 164970 J - T* 172.8 J

-164970 J > - T* 172.8 J

954.7< T

The system makes the transition from nonspontaneous to spontaneous at a temperature of 954.7 K.

Under 954.7 K the reaction is nonspontaneous; more than 954.7 K is the reaction spontaneous.

8 0
3 years ago
How did the lab activities help you answer the lesson question, "How are minerals and rocks classified?” What did you learn from
lorasvet [3.4K]

Answer:

In the lab, students were encouraged to look at different types of rocks and minerals. They were also told to observe the physical properties of the rocks and minerals. Based on these physical properties, the children were able to distinguish the different types of rocks and minerals. The major differences which the children will focus on will be the color, shape, hardness, luster, streak, fracture and cleavage of the rocks and minerals.

6 0
3 years ago
Read 2 more answers
Suggest why sodium wasn’t produced until the early 19th century.
AlexFokin [52]

Answer:

The reason is that sodium attaches itself very strongly to other elements. Its compounds are very difficult to break apartand also because it is so reactive.

5 0
3 years ago
How many moles of na2co3 are necessary to reach stoichiometric quantities with cacl2
lbvjy [14]

0.0102 moles Na₂CO₃ = 1.08g of Na₂CO₃ is necessary  to reach stoichiometric quantities with cacl2.

<h3>Explanation:</h3>

Based on the reaction

CaCl₂ + Na₂CO₃ → 2NaCl + CaCO₃

1 mole of CaCl₂ reacts per mole of Na₂CO₃

we have to calculate how many moles of CaCl2•2H2O are present in 1.50 g

  • We must calculate the moles of CaCl2•2H2O using its molar mass (147.0146g/mol) in order to answer this issue.
  • These moles, which are equal to moles of CaCl2 and moles of Na2CO3, are required to obtain stoichiometric amounts.
  • Then, we must use the molar mass of Na2CO3 (105.99g/mol) to determine the mass:

<h3>Moles CaCl₂.2H₂O:</h3>

1.50g * (1mol / 147.0146g) = 0.0102 moles CaCl₂.2H₂O = 0.0102moles CaCl₂

Moles Na₂CO₃:

0.0102 moles Na₂CO₃

Mass Na₂CO₃:

0.0102 moles * (105.99g / mol) = 1.08g of Na₂CO₃ are present

Therefore, we can conclude that 0.0102 moles Na₂CO₃  is necessary.to reach stoichiometric quantities with cacl2.

To learn more about stoichiometric quantities visit:

<h3>brainly.com/question/28174111</h3>

#SPJ4

7 0
2 years ago
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