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3 years ago

A blank is a condition that strays from normal homeostasis.

Chemistry

1 answer:

AURORKA [14]3 years ago

8 0

homeostatic imbalance is the answer, because it's when the internal environment cannot remain in equilibrium.

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Calculate the enthalpy change, ∆H in kJ, for the reaction H2O(s) → H2(g) + 1/2O2(g). Use the following information: : +279.9 kJ

Len [333]

Answer:

+ 291.9 kJ

Solution:

The equation given is as;

H₂O ₍s₎ → H₂ ₍g₎ + 1/2 O₂ ₍g₎ ΔH = ?

First, as we know the heat of formation of H₂O ₍l₎ is,

H₂ ₍g₎ + 1/2 O₂ ₍g₎ → H₂O ₍l₎ ΔH = - 285.9 kJ

Now, reversing the equation will reverse the sign of heat as,

H₂O ₍l₎ → H₂ ₍g₎ + 1/2 O₂ ₍g₎ ΔH = + 285.9 kJ

Also, we know that,

H₂O ₍s₎ → H₂O ₍l) ΔH = + 6.0 kJ

Now, adding last two equations,

H₂O ₍l₎ → H₂ ₍g₎ + 1/2 O₂ ₍g₎ ΔH = + 285.9 kJ

H₂O ₍s₎ → H₂O ₍l) ΔH = + 6.0 kJ

-----------------------------------------------------------------------------

H₂O ₍s₎ → H₂ ₍g₎ + 1/2 O₂ ₍g₎ ΔH = + 291.9 kJ

5 0

2 years ago

For the reaction 2 H2S(g) D 2 H2 (g) + S2 (g), Kp = 1.5 × 10−5 at 800.0°C. If the initial partial pressures of H2 and S2 in a cl

Usimov [2.4K]

Answer: The approximate equilibrium partial pressure of $H_2S$ is 3.92 atm

Explanation:

Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.

The given balanced equilibrium reaction is,

$2H_2S(g)\rightleftharpoons 2H_2(g)+S_2(g)$

$K_p=\frac{[H_2]^2\times [S_2]}{[H_2S]^2}$

$1.5\times 10^{-5}=\frac{[H_2]^2\times [S_2]}{[H_2S]^2}$

On reversing the reaction:

$2H_2(g)+S_2(g)\rightleftharpoons 2H_2S(g)$

initial pressure 4.00atm 2.00 atm 0

eqm (4.00-2x)atm (2.00-x) atm 2x atm

$K_p=\frac{[H_2S]^2}{[H_2]^2\times [S_2]}$

$K_p'=\frac{1}{K_p}=0.67\times 10^5$

$2H_2(g)+S_2(g)\rightleftharpoons 2H_2S(g)$

$0.67\times 10^5=\frac{2x]^2}{[4.00-2x]^2\times [2.00-x]}$

$x=1.96$

$[H_2S]=2x=2\times 1.96=3.92 atm$

Thus approximate equilibrium partial pressure of $H_2S$ is 3.92 atm

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