Answer:
The current temperature is -15° (15° below zero)
Explanation:
The temperature drops 10°:
T-10°
It will reach 25° below zero:
T - 10° = -25°
We add 10° in both members of the equation:
T - 10° +10° = -25° +10°
The equation is simplified as follows:
T = -25° +10°
T = -15°
The integer -15° can be expressed as 15° below zero.
Answer:
Ka = 1.52 E-5
Explanation:
- CH3-(CH2)2-COOH ↔ CH3(CH2)2COO- + H3O+
⇒ Ka = [H3O+][CH3)CH2)2COO-] / [CH3(CH2)2COOH]
mass balance:
⇒<em> C</em> CH3(CH2)2COOH = [CH3(CH2)2COO-] + [CH3(CH2)2COOH] = 1.0 M
charge balance:
⇒ [H3O+] = [CH3(CH2)2COO-]
⇒ Ka = [H3O+]²/(1 - [H3O+])
∴ pH = 2.41 = - Log [H3O+]
⇒ [H3O+] = 3.89 E-3 M
⇒ Ka = (3.89 E-3)² / ( 1 - 3.89 E-3 )
⇒ Ka = 1.519 E-5
To solve this problem, we assume ideal gas so that we can
use the formula:
PV = nRT
since the volume of the flask is constant and R is
universal gas constant, so we can say:
n1 T1 / P1 = n2 T2 / P2
1.9 mol * (21 + 273 K) / 697 mm Hg = n2 * (26 + 273 K) /
841 mm Hg
<span>n2 = 2.25 moles</span>
Because there is only one stable ionic compound made up of potassium and chlorine, and that is KCl. So calling is "mono chloride" or similar would be redundant assuming you understand basic chemistry (i.e. knowing oxidation numbers of K is +1 and Cl is -1). When compounds can exist in multiple forms in nature like CO and CO2 you will preferably indicate it through the nomenclature, calling one a monoxide and the other a dioxide.
