Answer:
Option 5. 179L
Explanation:
The reaction is:
2 HCl + Na₂CO₃ → 2NaCl + H₂O + CO₂
The amount of collected CO₂ is 8 moles.
We apply the Ideal Gases Law at STP
STP are 1 atm of pressure and 273K of T°
P . V = n . R . T
1atm . V = 8 mol .0.082L.atm/mol.K . 273K
V = (8 mol .0.082L.atm/mol.K . 273K) / 1 atm = 179 L
Answer:
1.33 Å
Explanation:
Given that the edge length , a of the KCl which forms the FCC lattice = 6.28 Å
Also,
For the FCC lattice in which the anion-cation contact along the cell edge , the ratio of the radius of the cation to that of anion is 0.731.
Thus,
.................1
Also, the sum of the radius of the cation and the anion in FCC is equal to half of the edge length.
Thus,
...................2
Given that:
To find,
Using 1 and 2 , we get:
<u>Size of the potassium ion = 1.33 Å</u>
Answer:
They all have the same fundamental properties of reflection
Answer:
2.0202 grams
Explanation:
1.4% (m/v) glucose solution means: 1.4g glucose/100mL solution.
so ?g glucose = 144.3 mL soln
Now apply the conversion factor, and you have:
?g glucose = 144.3mL soln x (1.4g glucose/100mL soln).
so you have (144.3x1.4/100) g glucose= 2.0202 grams
The amount of the solute is constant during dilution. So the mole number of HCl is 2*1.5=3 mole. The volume of HCl stock is 3/12=0.25 L. So using 0.25 L stock solution and dilute to 2.0 L.
