Answer:
-0.1767°C (Option A)
Explanation:
Let's apply the colligative property of freezing point depression.
ΔT = Kf . m. i
i = Van't Hoff factot (number of ions dissolved). Glucose is non electrolytic so i = 1
m = molality (mol of solute / 1kg of solvent)
We have this data → 0.095 m
Kf is the freezing-point-depression constantm 1.86 °C/m, for water
ΔT = T° frezzing pure solvent - T° freezing solution
(0° - T° freezing solution) = 1.86 °C/m . 0.095 m . 1
T° freezing solution = - 1.86 °C/m . 0.095 m . 1 → -0.1767°C
Answer:
3.43 %
Explanation:
We need to calculate first the number of moles of CeO2 produced in the combustion. Given its formula we know how many moles of Ce atom are present. From there calculate the mass this number of moles this represent and then one can calculate the percentage.
0.1848 g CeO2 x 1 mol CeO2/172.114g = 0.00107 mol CeO2
0.00107 mol CeO2 x 1 mol Ce/ 1 mol CeO2 = 0.00107 mol Ce
.00107 mol Ce x 140.116 g Ce/ mol = 0.150 g Ce
0.150 g Ce/ 4.3718 g sample x 100 = 3.43 %
Selfmade.ivyy hope this help
Answer:
Your answer is Gravity,
The Earth's gravity pulls things at a rate of 9.8m/s² (varying depending upon one's distance). according to Newton, "all mass in the universe exerts some gravitational force". you are pulling on Sagitarios A* (the black hole at the center of the milky way galaxy) right now, and it is also pulling on you.
If you have any questions I <em><u>insist</u></em> that you ask in the comments.
<em>Wbob1314</em>
