To plan ahead using REPS and Checks, choose the minimum space cushion needed to maintain folloying distance traveling at 25 MPH

PART A: Use the following glycolytic reaction to answer the question. If the concentration of DHAP is 0.125 M and the concentrat

alisha [2963]

Answer:

For A: The change in free energy for the reaction is -5339.76 J/mol

For B: Free energy change is expressed in kJ/mol

For C: The forward reaction favors progression, while the reverse reaction does not.

Explanation:

Regarding the specified chemical reaction:

$DHAP\rightleftharpoons G_3P$

For A:

The relationship between standard Gibbs free energy and equilibrium constant is as follows:

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

The free energy change can be calculated using the following equation:

$\Delta G=\Delta G^o+RT\ln Q$

Or,

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

where,

$\Delta G$ = Change in free energy

R = Gas constant = $8.314J/K mol$

$T^o$ = standard temperature = $25^oC=[273+25]K=298K$

T = temperature of the cell = $37^oC=[273+37]K=310K$

$K_[eq}$ = equilibrium constant = $5.4\times 10^{-2}$

Q = reaction quotient = $\frac{[G_3P]}{[DHAP]}$

$[G_3P]$ = 0.06 M

[DHAP] = 0.125 M

Substituting the values into the equation yields:

$\Delta G=[-(8.314J/mol.K\times 298K\times \ln (5.4\times 10^{-2}))]+[(8.314J/mol.K\times 310K\times \ln (\frac{0.06}{0.125}))]\\\\\Delta G=-[-7231.46]+[-1891.7]=-5339.76J/mol$

Thus, the change in free energy for the reaction is -5339.76 J/mol

For B:

To convert the free energy change to kilojoules, we apply the conversion factor:

1 kJ = 1000 J

So, $-5339.76J/mol\times \frac{1kJ}{1000J}=-5.34kJ/mol$

Consequently, the free energy change's units are kJ/mol

For C:

For spontaneity in the reaction, the Gibbs free energy must be negative. However, the calculations indicate a positive Gibbs free energy, leading to the conclusion that the reaction is not spontaneous.

The free energy change of the reaction is negative.

Consequently, the forward reaction is favored and the reverse reaction is not favored.

8 0

3 months ago

If 32.0 g of MgSO4⋅7H2O is thoroughly heated, what mass of anhydrous magnesium sulfate will remain?

alisha [2963]

To calculate the moles of MgSO4.7H2O, we find the molar mass equals 246, thus moles = 32 / 246 = 0.13 moles. Upon heating, all 7 H2O from one molecule will evaporate. The total moles of H2O present amount to 7 x 0.13 = 0.91, and the mass of that H2O is 0.91 x 18 = 16.38g. Therefore, the mass of the anhydrous MgSO4 that remains is 32 - 16.38 = 15.62 g.

6 0

2 months ago

How many micrograms of iron were in the 8.0 mL sample of Greg's blood?

castortr0y [3046]

The result is: 3.36 micrograms of iron in<span> Greg's blood sample.
</span>m(Fe) = 42 mcg(micrograms).
V(Fe) = 1 dL = 1 dL · 100 mL/1dL.
V(Fe) = 100 mL.
Using proportions: m(Fe): 8 mL = 42 mcg: 100 mL.
Thus, 100 mL · m(Fe) = 8 mL · 42 mcg.
m(Fe) = 336 mL·mcg ÷ 100 mL.
m(Fe) = 3.36 mcg.

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Suppose you held a lighted match to a solid hunk of wood and another match to a pile of wood shavings. Which form of wood will c

Tems11 [2777]

Answer:

Wood

Explanation:

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The process of water migrating from areas of low electrolyte concentration to area of high electrolyte concentration is called —

alisha [2963]

1-Diffusion

2-Cerebral edema

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