Mass 1 + %abundance of first isotope + Mass 2 + %abundance of second isotope
/ 100
This is RAM.
Answer:
1528.3L
Explanation:
To solve this problem we should know this formula:
V₁ / T₁ = V₂ / T₂
We must convert the values of T° to Absolute T° (T° in K)
21°C + 273 = 294K
70°C + 273 = 343K
Now we can replace the data
1310L / 294K = V₂ / 343K
V₂ = (1310L / 294K) . 343K → 1528.3L
If the pressure keeps on constant, volume is modified directly proportional to absolute temperature. As T° has increased, the volume increased too
H2SO3 or sulfurous acid is actually a strong acid. We know
for a fact that strong acids completely dissociate into its component ions in a
solution, that is:
<span>H2SO3 --> 2H+ + SO3-</span>
<span>So from the equation above, there are 2 moles of H+</span>
Answer:
1,15mL = V₂
Explanation:
Based on Charle's law the volume is directely proportional to the absolute temperature in a gas under constant pressure. The equation is:
V₁T₂ = V₂T₁
<em>Where V is volume and T absolute temperature of a gas where 1 is initial state and 2, final state.</em>
The V₁ is 1.23mL
T₁ = 32°C + 273.15 = 305.15K
T₂ = T₁ - 20°C = 285.15K
Replacing:
1.23mL*285.15K = V₂*305.15K
<h3>1,15mL = V₂</h3>
<em />
Answer:
0.200 m K3PO3
Explanation:
Let us remember that the freezing point depression is obtained from the formula;
ΔTf = Kf m i
Where;
Kf = freezing point constant
m = molality
i = Van't Hoff factor
The Van't Hoff factor has to do with the number of particles in solution. Let us consider the Van't Hoff factor for each specie.
0.200 m HOCH2CH2OH - 1
0.200 m Ba(NO3)2 - 3
0.200 m K3PO3 - 4
0.200 m Ca(CIO4)2 - 3
Hence, 0.200 m K3PO3 has the greatest van't Hoff factor and consequently the greatest freezing point depression.
