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2 years ago

5

Explain how our body systems work together to get oxygen into and around our body.

1 answer:

aleksandrvk [35]2 years ago

7 0

Answer: (1) Inhales (breathes in) Oxygen - good for the body - gives it to the Circulatory System to be transported throughout the body through the blood. (1) Digestive System gets nutrients (good) from food and hands it over to the blood and Circulatory System then carries those nutrients where they need to go.

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What does morality measure

allsm [11]

Morality represents a society's positive value orientation, providing a basis for the assessment of the appropriateness of social behavior. Moral behavior in our model is not subject to regional and cultural influences

6 0

2 years ago

What elements were formed in Nuclear fusion?

Sonja [21]

Answer:

tritium and deuterium are combined and result in the formation of helium

6 0

3 years ago

n acid with a pKa of 8.0 is present in a solution with a pH of 6.0. What is the ratio of the protonated to the deprotonated form

Nady [450]

Answer : The ratio of the protonated to the deprotonated form of the acid is, 100

Explanation : Given,

$pK_a=8.0$

pH = 6.0

To calculate the ratio of the protonated to the deprotonated form of the acid we are using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

$pH=pK_a+\log \frac{[Deprotonated]}{[Protonated]}$

Now put all the given values in this expression, we get:

$6.0=8.0+\log \frac{[Deprotonated]}{[Protonated]}$

$\frac{[Deprotonated]}{[Protonated]}=0.01$

As per question, the ratio of the protonated to the deprotonated form of the acid will be:

$\frac{[Protonated]}{[Deprotonated]}=100$

Therefore, the ratio of the protonated to the deprotonated form of the acid is, 100

5 0

2 years ago

In which group of the Periodic Table do most of the elements exhibit both positive and negative oxidation states?

Shtirlitz [24]

C i think but you should pick it anyway

4 0

3 years ago

For the reaction C2H4(g) + H2O(g) --> CH3CH2OH(g)

Dominik [7]

Answer : The value of equilibrium constant for this reaction at 262.0 K is $3.35\times 10^{2}$

Explanation :

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

$\Delta H^o$ = standard enthalpy = -45.6 kJ = -45600 J

$\Delta S^o$ = standard entropy = -125.7 J/K

T = temperature of reaction = 262.0 K

Now put all the given values in the above formula, we get:

$\Delta G^o=(-45600J)-(262.0K\times -125.7J/K)$

$\Delta G^o=-12666.6J=-12.7kJ$

The relation between the equilibrium constant and standard Gibbs free energy is:

$\Delta G^o=-RT\times \ln k$

where,

$\Delta G^o$ = standard Gibbs free energy = -12666.6 J

R = gas constant = 8.314 J/K.mol

T = temperature = 262.0 K

K = equilibrium constant = ?

Now put all the given values in the above formula, we get:

$-12666.6J=-(8.314J/K.mol)\times (262.0K)\times \ln k$

$k=3.35\times 10^{2}$

Therefore, the value of equilibrium constant for this reaction at 262.0 K is $3.35\times 10^{2}$

3 0

3 years ago