What is formed when a BASE dissociates in water, according to the Brønsted-Lowry definition?

in their mechanism of action, a difference between lipid-soluble and water-soluble hormones is that _____.

LenKa [72]

In their mechanism of action, a difference between lipid-soluble and water-soluble hormones is that lipid-soluble hormones diffuse into the cell's plasma membrane to access protein receptors, while water-soluble hormones bind to protein receptors on the plasma membrane without diffusion.

Examples of lipid-soluble hormones are steroids. Examples of water-soluble hormones are polypeptides.

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4 years ago

8. A sample of potassium chlorate (KCIO,) was heated in a test tube and decomposed 2KC?(s) 302 (g) + 2KCIO, (s) The oxygen was c

dangina [55]

Answer:

Partial pressure of $O_{2}$ in the gas was 733 torr and mass of $KClO_{3}$ in the sample was 2.12 g.

Explanation:

a) Total pressure of gas = (partial pressure of water vapour)+(partial pressure of $O_{2}$ )

Here partial pressure of water vapour is 21 torr and total pressure of gas is 754 torr.

So, partial pressure of $O_{2}$ = (total pressure of gas)-(partial pressure of water vapour) = (754 torr) - (21 torr) = 733 torr

b) Lets assume that $O_{2}$ behaves ideally. Hence-

PV=nRT

where P is pressure of $O_{2}$ , V is volume of $O_{2}$ , n is number of moles of $O_{2}$ , R is gas constant and T is temperature in kelvin

here P = 733 torr = $(733\times 0.001316)atm$ = 0.9646 atm

V = 0.65 L, R = 0.082 L.atm/(mol.K), T=(273+22)K = 295 K

So, $n=\frac{PV}{RT}$

= $\frac{(0.9646 atm)\times (0.65 L)}{(0.082 L.atm/(mol.K))\times (295 K)}$

= 0.0259 moles

As 3 moles of $O_{2}$ are produced from 2 moles of $KClO_{3}$ therefore 0.0259 moles of $O_{2}$ are produced from $(\frac{2\times 0.0259}{3})$ moles or 0.0173 moles of $KClO_{3}$ .