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Brut [27]
2 years ago
13

Which of the reactions are spontaneous (favorable)? DHAP − ⇀ ↽ − glyceraldehyde-3-phosphate Δ G = 3.8 kJ / mol DHAP↽−−⇀glycerald

ehyde-3-phosphateΔG=3.8 kJ/mol C 4 H 4 O 5 ⟶ C 4 H 2 O 4 + H 2 O Δ G = 3.1 kJ / mol C4H4O5⟶C4H2O4+H2OΔG=3.1 kJ/mol C 2 H 4 + H 2 Rh ( I ) −−−→ C 2 H 6 Δ G = − 150.97 kJ / mol C2H4+H2→Rh(I)C2H6ΔG=−150.97 kJ/mol glutamate + NAD + + H 2 O ⟶ NH + 4 + α-ketoglutarate + NADH + H + Δ G = 3.7 kcal / mol glutamate+NAD++H2O⟶NH4++α-ketoglutarate+NADH+H+ΔG=3.7 kcal/mol L -malate + NAD + ⟶ oxaloacetate + NADH + H + Δ G = 29.7 kJ / mol -malate+NAD+⟶oxaloacetate+NADH+H+ΔG=29.7 kJ/mol C 6 H 13 O 9 P + ATP ⟶ C 6 H 14 O 12 P 2 + ADP Δ G = − 14.2 kJ / mol
Chemistry
1 answer:
Tpy6a [65]2 years ago
7 0

Answer:

Explanation:

In spontaneous reaction , there is decrease in Gibb's free energy .( Δ G is negative ). Out of given reaction , following reactions have negative Δ G so they are spontaneous.

C ₂ H ₄ + H ₂ Rh ( I ) −−−→ C ₂ H ₆ ,  Δ G = − 150.97 kJ / mol

C ₆ H₁₃O₉ P + ATP ⟶ C ₆ H₁₄ O₁₂ P₂ + ADP ,  Δ G = − 14.2 kJ / mol

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Calculate the molarity of 1.0 mol of KCl in 750 ml of solution
Vaselesa [24]

Answer:

[KCl] = 1.33 M

Explanation:

Molarity is mol /L

Mol of solute in 1 L of solution

Volume of solution is 750 mL

750 mL / 1000 = 0.750 L

1 mol / 0.750L = M → 1.33

7 0
2 years ago
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Ammonia gas occupies a volume of 2,725ml at a pressure of 701 kPa. What volume would it occupy at 101 kPa?
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A solution is made by dissolving 9.74 g of sodium sulfate in water to a final volume of 165 mL of solution. What is the weight/w
MatroZZZ [7]

Answer: The weight/weight % or percent by mass of the solute is 5.41 %.

Explanation:

Mass of the sodium sulfate,w = 9.74 g

Volume of the water = 165 mL

Density of the water = 1 g/mL

Density=1 g/mL=\frac{\text{Mass of water}}{\text{Volume of water}}

Mass of the water =1 g/mL\times 165 mL=165 g

Mass of the solution, W:

Mass of solute + Mass of solvent =9.47 g + 165 g=174.47 g

w/w\%=\frac{w\times 100}{W}=\frac{9.45 g\times 100}{174.47 g}=5.41 \%

The weight/weight % or percent by mass of the solute is 5.41 %.


8 0
2 years ago
Calculate the heat change in calories for melting 65 g of ice at 0 ∘c.
Genrish500 [490]

When ice melts, the physicals state changes from solid to liquid. The energy or the heat required (q) required to change a unit mass (m) of a substance from solid to liquid is known as the enthalpy or heat of fusion (ΔHf). The variables; q, m and ΔHf are related as:

q = m * ΔHf

the mass of ice m = 65 g

the heat of fusion of water at 0C = ΔHf = 334 J/g

Therefore: q = 65 g * 334 J/g = 21710 J

Now:

4.184 J = 1 cal

which implies that: 21710 J = 1 cal * 21710 J/4.184 J = 5188.8 cal

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5 0
2 years ago
A 23.2 g sample of an organic compound containing carbon, hydrogen and oxygen was burned in excess oxygen and yielded 52.8 g of
allochka39001 [22]

Answer:

The answer to your question is   C₃H₆O

Explanation:

Data

mass of sample = 23.2 g

mass of carbon dioxide = 52.8 g

mass of water = 21.6 g

empirical formula = ?

Process

1.- Calculate the mass and moles of carbon

                       44 g of CO₂ ---------------  12 g of C

                        52.8 g          ---------------  x

                        x = (52.8 x 12)/44

                        x = 633.6/44

                        x = 14.4 g of C

                        12 g of C ------------------  1 mol

                        14.4 g of C ---------------   x

                         x = (14.4 x 1)/(12)

                         x = 1.2 moles of C

2.- Calculate the grams and moles of Hydrogen

                         18 g of H₂O ---------------  2 g of H

                         21.6 g of H₂O -------------  x

                          x = (21.6 x 2) / 18

                         x = 2.4 g of H

                         1 g of H -------------------- 1 mol of H

                         2.4 g of H -----------------  x

                          x = (2.4 x 1)/1

                          x = 2.4 moles of H

3.- Calculate the grams and moles of Oxygen

Mass of Oxygen = 23.2 - 14.4 - 2.4

                           = 6.4 g

                         16 g of O ----------------  1 mol

                          6.4 g of O --------------  x

                          x = (6.4 x 1)/16

                          x = 0.4 moles of Oxygen

4.- Divide by the lowest number of moles

Carbon = 1.2 / 0.4 = 3

Hydrogen = 2.4/ 0.4 = 6

Oxygen = 0.4 / 0.4 = 1

5.- Write the empirical formula

                                C₃H₆O

8 0
2 years ago
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