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3 years ago

11

Help me with this please !!!

1 answer:

Daniel [21]3 years ago

4 0

militer is the answer

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If a compound has a log kow value of 6.5, what would be its predicted concentration (in ppm) in the fat of fish that swim in wat

blsea [12.9K]

Easy ans I don't know .......

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4 years ago

What is the name of the force exerted by blood on the walls of arteries

Digiron [165]

The name of the force exerted by blood on the wall of the arteries is blood pressure. Blood<span> pressure is the </span>force<span> of </span>blood<span> against the </span>walls<span> of the </span>arteries<span>.
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3 years ago

Pertain to the following class, Point:

Anton [14]

Answer:

Explanation:

b = 0;

this (x, y);

this(0, 0);

x = a;

y = b;

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5 0

3 years ago

Methane, a component in natural gas, can be used as a fuel in combustion reactions. What is the value for ΔGnon (in kJ) for the

Oksanka [162]

<u>Answer:</u> The $\Delta G$ for the reaction is -806.86 kJ

<u>Explanation:</u>

We are given:

$\Delta H^o_{rxn}=-803kJ=-803000J$ (Conversion factor: 1 kJ = 1000)

$\Delta S^o_{rxn}=-4.05J/K$

Temperature of the reaction = 293 K

To calculate the standard Gibbs's free energy of the reaction, we use the equation:

$\Delta G^o_{rxn}=\Delat H^o_{rxn}-T\Delta S^o_{rxn}$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=-803000J-[(293K)\times (-4.05J/K)]=-801813.35J$

For the given chemical equation:

$CH_4(g)+2O_2(g)\rightleftharpoons CO_2(g)+2H_2O(g)$

The expression for $K_c$ is given as:

$K_{c}=\frac{[H_2O]^2[CO_2]}{[CH_4][O_2]^2}$

We are given:

$[H_2O]=6.41M$

$[CO_2]=3.83M$

$[CH_4]=14.51M$

$[O_2]=9.27M$

Putting values in above equation, we get:

$K_c=\frac{(6.41)^2\times 3.83}{14.51\times (9.27)^2}$

$K_c=0.126$

To calculate the Gibbs free energy of the reaction, we use the equation:

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

where,

$\Delta G$ = Gibbs' free energy of the reaction = ?

$\Delta G^o$ = Standard gibbs' free energy change of the reaction = -801813.35 J

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

T = Temperature = 293 K

$K_c$ = equilibrium constant in terms of concentration = 0.126

Putting values in above equation, we get:

$\Delta G=-801813.35J+(8.314J/K.mol\times 293K\times \ln(0.126))$

$\Delta G=-806859.46J=-806.86kJ$

Hence, the $\Delta G$ for the reaction is -806.86 kJ

8 0

3 years ago

Why should we work to protect estuaries? Select all that apply.<br> PLSSSS

True [87]

Answer:

1, 3, and 4

Explanation:

I'm guessing, but those make the most sense

8 0

3 years ago

Read 2 more answers