ِAnswer:
1- The molarity of HCOOH = 9.515 M.
2- The mole fraction of HCOOH = 0.18.
Explanation:
<em>1- The molarity of HCOOH:</em>
- We can calculate the molarity of HCOOH using the relation:
M = (10pd)/molar mass.
p is the percent by mass of HCOOH = 35.9 %.
d is the specific gravity of HCOOH = 1.22 g/cm³.
Molar mass of HCOOH = 46.03 g/mol.
∴ M = (10pd)/molar mass = (10)(35.9 %)(1.22 gcm³) / (46.03 g/mol) = 9.515 M.
<em>2- The mole fraction of HCOOH:</em>
- We can suppose that we have a 100 g solution, that contains 35.9 g of HCOOH and 64.1 g of water.
<em>The mole fraction of HCOOH = (no. of moles of HCOOH) / (no. of moles of HCOOH + no, of moles of water).</em>
no. of moles of HCOOH = mass / molar mass = (35.9 g)/(46.03 g/mol) = 0.78 mol.
no. of moles of water = mass / molar mass = (64.1 g)/(18.0 g/mol) = 3.56 mol.
- The mole fraction of HCOOH = (no. of moles of HCOOH) / (no. of moles of HCOOH + no, of moles of water) = (0.78 mol) / (0.78 mol + 3.56 mol) = 0.18.
The flame that comes out of the Bunsen burner. it's blue/pale violet.
This compound is also known as Calcium Oxide. The SI base unit for amount of substance is the mole. 1 grams CaO is equal to 0.017832495800447 mole. Note that rounding errors may occur, so always check the results.
Hope I helped :)
The fraction of Earth's radius (6371 km) relative to the thickness of the oceanic (7.5 km) and continental crust (35 km) is 0.12 and 0.55, respectively.
What we know:
- The average radius of Earth (E) = 6371 km
- The average thickness of oceanic crust (O) = 7.5 km
- The average thickness of continental crust (C) = 35 km
We need to convert all the above units from kilometers to miles:
Now, we can calculate the fraction of Earth's radius relative to each type of crust, with the given equation:
- <u>For the oceanic crust (O)</u>:
- <u>For the continental crust (C)</u>:
Therefore, the fraction of Earth's radius relative to the oceanic and continental crust is 0.12 and 0.55, respectively.
You can see another example of calculation of fractions of Earth's radius here: brainly.com/question/4675868?referrer=searchResults
I hope it helps you!
Answer:
Oxidation occurs at the anode: Fe(s) | Fe2+(aq) half cell. (ii) Reduction occurs at the cathode: Ag(s) | Ag+(aq) half cell. Oxidation occurs at the anode: Pt | Sn2+(aq), Sn4+(aq) half cell. (iii) Electrons flow from the anode to the cathode: from the Pt(s) → Ag(s) electrode.
