The prefix for one-millionth is micro-. True False

Consider the malate dehydrogenase reaction from the citric acid cycle. Given the listed concentrations, calculate the free energ

Ahat [919]

Answer: The Gibbs free energy of the reaction is 21.32 kJ/mol

Explanation:

The chemical equation follows:

$\text{Malate }+NAD^+\rightleftharpoons \text{Oxaloacetate }+NADH$

The equation used to Gibbs free energy of the reaction follows:

$\Delta G=\Delta G^o+RT\ln K_{eq}$

where,

$\Delta G$ = free energy of the reaction

$\Delta G^o$ = standard Gibbs free energy = 29.7 kJ/mol = 29700 J/mol (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = 8.314J/K mol

T = Temperature = $37^oC=[273+37]K=310K$

$K_{eq}$ = Ratio of concentration of products and reactants = $\frac{\text{[Oxaloacetate]}[NADH]}{\text{[Malate]}[NAD^+]}$

$\text{[Oxaloacetate]}=0.130mM$

$[NADH]=2.0\times 10^2mM$

$\text{[Malate]}=1.37mM$

$[NAD^+]=490mM$

Putting values in above expression, we get:

$\Delta G=29700J/mol+(8.314J/K.mol\times 310K\times \ln (\frac{0.130\times 2.0\times 10^2}{1.37\times 490}))\\\\\Delta G=21320.7J/mol=21.32kJ/mol$

Hence, the Gibbs free energy of the reaction is 21.32 kJ/mol

4 0

3 years ago

a scientist uses 68 grams of CaCo3 to prepare 1.5 liters of solution. what is the molarity of this solution?

victus00 [196]

Answer: 0.4533mol/L

Explanation:

Molar Mass of CaCO3 = 40+12+(16x3) = 40+12+48 = 100g/mol

68g of CaCO3 dissolves in 1.5L of solution.

Xg of CaCO3 will dissolve in 1L i.e

Xg of CaCO3 = 68/1.5 = 45.33g/L

Molarity = Mass conc.(g/L) / molar Mass

Molarity = 45.33/100 = 0.4533mol/L

7 0

4 years ago

A sample of propane (C3H8) is placed in a closed vessel together with an amount of O2 that is 3 times the amount needed to compl

Aneli [31]

Answer:

final mole fraction of O₂ = 58.84% , CO₂ = 17.64% , H₂O = 23.52% .

final partial pressure of O₂ = 2.942 atm , CO₂ = 0.882 atm , H₂O = 1.176 atm .

Explanation:

Assuming that propane is present as a gas , and also that the combustion is complete:

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

then taking as a reference propane sample= 1 mol , then

initial moles of O₂ = 3* 5 moles = 15 moles

final moles of O₂= 3* 5 moles - 5 moles = 10 moles

final moles of CO₂ = 3 moles

final moles of H₂O = 4 moles

total number of moles = 10 + 3 + 4 = 17 moles

final mole fraction of O₂ = 10/17 = 0.5884 = 58.84%

final mole fraction of CO₂ = 3/17 = 0.1764 = 17.64%

final mole fraction of H₂O = 4/17 = 0.2352 = 23.52%

From Dalton's law for ideal gases , the partial pressure p=P*x then

final partial pressure of O₂ = 5 atm * 10/17 = 2.942 atm

final partial pressure of CO₂ = 5 atm * 3/17 = 0.882 atm

final partial pressure of H₂O = 5 atm * 1/17 = 1.176 atm

4 0

3 years ago

Read 2 more answers

choose a spot outside or inside your house as your reference point. it may be a trer ,bench,cabinet,refrigerator,pr sink.

fomenos

Answer:

I choose my table

6 0

3 years ago

Ethanol (c2h5oh) melts at -114c and boils at 78 c the enthalpy of fusion of ethanol is 5.02 kj/mol, and its enthalpy of vaporiza

vovikov84 [41]

100 kilo joules There are several phases that this problem undergoes and the final answer is the sum of all the energy used for each phase. Phase 1. Heating of solid ethanol until its melting point. Phase 2. Melting of the ethanol until it's completely liquid. Phase 3. Heating of the liquid ethanol until it reaches its boiling point. Phase 4. Boiling the ethanol until it's completely vapor. To make things more interesting, some of our constant are per gram and some others are per mole. So let's calculate how many moles of ethanol we have. Atomic weight carbon = 12.0107 Atomic weight hydrogen = 1.00794 Atomic weight oxygen = 15.999 Molar mass ethanol = 2*12.0107 + 6*1.00794 + 15.999 = 46.06804 g/mol Moles ethanol = 75g / 46.06804 g/mol = 1.628026719 mol Phase 1. Use the specific heat of solid ethanol and multiply by the number of degrees we need to change by the mass we have. So 0.97 J/g*K * 75 g * (-114c - -120c) = 0.97 J/g*K * 75 g * 6K = 436.5 J Phase 2: Time to melt. Just need the moles and the enthalpy of fusion. So: 1.628026719 mol * 5.02 kJ/mol = 8.172694128 kJ Phase 3: Heat to boiling. Just like heating to melting, just a different specific heat and temperature 2.3J/g*K * 75g * (78c - -114c) = 2.3J/g*K * 75g * 192 K = 33120 J Phase 4: Boil it to vapor. Need moles and enthalpy of vaporization. So 1.628026719 mol * 38.56 kJ/mol = 62.77671027 kJ Now let's add them together: 436.5 J + 8.172694128 kJ + 33120 J + 62.77671027 kJ = 0.4365 kJ + 8.172694128 kJ + 33.120 kJ + 62.77671027 kJ =104.5059044 kJ Since the least precise datum we have is 2 significant figures, round the result to 2 significant figures, giving 100 kilo joules.

3 0

4 years ago