Answer:
I guess A, I am not sure...
Answer:
148.04 kJ/mol
Explanation:
Let's consider the following thermochemical equation.
NO(g) + 1/2 O₂(g) → NO₂(g) ΔH°rxn = -114.14 kJ/mol
We can find the standard enthalpy of formation (ΔH°f) of NO(g) using the following expression.
ΔH°rxn = 1 mol × ΔH°f(NO₂(g)) - 1 mol × ΔH°f(NO(g)) - 1/2 mol × ΔH°f(O₂(g))
ΔH°f(NO(g)) = 1 mol × ΔH°f(NO₂(g)) - ΔH°rxn - 1/2 mol × ΔH°f(O₂(g)) / 1 mol
ΔH°f(NO(g)) = 1 mol × 33.90 kJ/mol - (-114.14 kJ) - 1/2 mol × 0 kJ/mol / 1 mol
ΔH°f(NO(g)) = 148.04 kJ/mol
When E° cell is an electrochemical cell which comprises of two half cells.
So,
when we have the balanced equation of this half cell :
Al3+(aq) + 3e- → Al(s) and E°1 = -1.66 V
and we have also this balanced equation of this half cell :
Ag+(aq) + e- → Ag(s) and E°2 = 0.8 V
so, we can get E° in Al(s) + 3Ag (aq) → Al3+(aq) + 3Ag(s)
when E° = E°2 - E°1
∴E° =0.8 - (-1.66)
= 2.46 V
∴ the correct answer is 2.46 V
Answer:
Solution A: crenation
Solution B: hemolysis
Solution C: hemolysis
Solution D: crenation
Solution E: crenation
Explanation:
Hemolysis is the rupturing of red blood cells, which results in the release of hemoglobin (from within the red blood cells) into the plasma. If a red blood cell is placed in a hypotonic solution, water will flow into the cell, the cell will swell and hemolysis will.
Crenation: when a red blood cell is placed in a <em>hypertonic solution (</em>such as highly saline solution), the red blood cell will lose water(osmosis) and will shrink in size. The red blood cell has undergone crenation.
In order for a red blood cell to prevent from undergoing hemolysis or crenation, the cell must be placed in an<em> isotonic solution, </em>i.e either in <u>0.9% (m/v) NaCl solution</u> or <u>5% glucose solution</u>
- Solution B and Solution C are hypotonic solution, thus red blood cell has undergone hemolysis.
- Solution A, D and E are hypertonic solution. thus red blood cell has undergone crenation
A thick liquid could be one property or its white.....
