The computation for this problem is:
(1.55x10^4 / 1.0x10^3) x 19.8 mm Hg
= 15.5 x 19.88 mm Hg
= 308.14 mm Hg decrease
= 308.14 x 0.05 C = 15.407 deg C
deduct this amount to 100
100 – 15.407 = 84.593 C
ANSWER: 85 deg C (rounded to 2 significant figures)
Answer:
Option A. KCl (aq)
Option D. Mg(OH)₂(s
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
MgCl₂(aq) + KOH(aq) —>
In solution, MgCl₂(aq) and KOH(aq) will dissociate as follow:
MgCl₂(aq) —> Mg²⁺(aq) + 2Cl¯(aq)
KOH(aq) —> K⁺(aq) + OH¯(aq)
MgCl₂(aq) + KOH(aq) —>
Mg²⁺(aq) + 2Cl¯(aq) + 2K⁺(aq) + OH¯(aq) —> 2K⁺(aq) + 2Cl¯(aq) + Mg(OH)₂ (s)
MgCl₂(aq) + KOH(aq) —> 2KCl (aq) + Mg(OH)₂(s)
Thus, the products of the above reaction are: KCl(aq) and Mg(OH)₂(s)
Thus, option A and D gives the correct answer to the question.
The electron geometry of a water molecule is tetrahedral even though the molecular geometry is bent.
As water molecule hybridisation is sp³ that provides it a electron geometry tetrahedral but due to presence of 2 lone pairs and 2 bond pairs its molecular geometry is bent.
The hybridisation sp³ makes electron geometry of a water molecule tetrahedral but the presence of 2 lone pairs makes its molecular geometry bent
Answer:
In a chemical equilibrium, the forward and reverse reactions occur at equal rates, and the concentrations of products and reactants remain constant. A catalyst speeds up the rate of a chemical reaction, but has no effect upon the equilibrium position for that reaction.
Explanation:
but heres a way to solve it
An athlete takes a deep breath, inhaling 1.85 L of air at 21°C and 754 mm Hg.
T
How many moles of air are in the breath? How many molecules?
Gas constant, R= 8.314 J mol ¹ K-1
PV = nRT
PV
RT
h=
=
P
= 0.08206 L atm mol-1 K-1
= 62.36 L Torr mol-1 K-1 -
1 atm = 760 mm Hg = 760 Torr
754 Forr 1.85€
6236 Jerr 294K
