Which of these would help maintain homeostasis during exercise?

Chemistry

2 answers:

a_sh-v [17]3 years ago

8 0

Answer:

D) Speeding up the rate of breathing

Explanation:

During homeostasis, the body increases breathing rate, heart rate, and sweat production to maintain homeostasis.

statuscvo [17]3 years ago

4 0

I believe the answer is d

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21+ Cl2 → 12+201 -

Alina [70]

Answer:

$\rm 2\; I^{-} + Cl_2 \to I_2 + 2 \; Cl^{-}$ .

Start color: yellowish-green.

End color: dark purple.

Assumption: no other ion in the solution is colored.

Explanation:

In this reaction, chlorine gas $\rm Cl_2$ oxidizes iodine ions $\rm I^{-}$ to elemental iodide $\rm I_2$ . At the same time, the chlorine atoms are converted to chloride ions $\rm Cl^{-}$ .

Fluorine, chlorine, bromine, and iodine are all halogens. They are all found in the 17th column of the periodic table from the left. One similarity is that their anions are not colored. However, their elemental forms are typically colored. Besides, moving down the halogen column, the color becomes darker for each element.

Among the reactants of this reaction, $\rm I^{-}$ is colorless. If there's no other colored ion, only the yellowish-green hue of $\rm Cl_2$ would be visible. Hence the initial color of the reaction would be the yellowish-green color of $\rm Cl_2$ .

Similarly, among the products of this reaction, $\rm Cl^{-}$ is colorless. If there's no other colored ion, only the dark purple hue of $\rm I_2$ would be visible. Hence the initial color of the reaction would be the dark purple color of $\rm I_2$ .

5 0

3 years ago

Nitric oxide reacts with chlorine to form nocl. the data refer to 298 k. 2no(g) + cl2(g) → 2nocl(g) substance: no(g) cl2(g) nocl

tigry1 [53]

Answer:

- 10.555 kJ/mol.

Explanation:

∵ ∆G°rxn = ∆H°rxn - T∆S°rxn.

Where, ∆G°rxn is the standard free energy change of the reaction (J/mol).

∆H°rxn is the standard enthalpy change of the reaction (J/mol).

T is the temperature of the reaction (K).

∆S°rxn is the standard entorpy change of the reaction (J/mol.K).

Calculating ∆H°rxn:

∵ ∆H°rxn = ∑∆H°products - ∑∆H°reactants

∴ ∆H°rxn = (2 x ∆H°f NOCl) - (1 x ∆H°f Cl₂) - (2 x ∆H°f NO) = (2 x 51.71 kJ/mol) - (1 x 0) - (2 x 90.29 kJ/mol) = - 77.16 kJ/mol.

Calculating ∆S°rxn:

∵ ∆S°rxn = ∑∆S°products - ∑∆S°reactants

∴ ∆S°rxn = (2 x ∆S° NOCl) - (1 x ∆S° Cl₂) - (2 x ∆S° NO) = (2 x 261.6 J/mol.K) - (1 x 223.0 J/mol.K) - (2 x 210.65 J/mol.K) = - 121.1 J/mol.K. = - 0.1211 kJ/mol.K.