Answer:
2C3H8O + 9O2 ==> 6CO2 + 8H2O ... balanced equation
moles propanol = 5.26 g x 1 mol/60.1 g = 0.0875 moles
moles O2 = 31.8 g x 1 mol/31.9 g = 0.997 moles O2
Propanol is limiting based on the mol ratio in balance equation of 2 : 9
To find mass of O2 (excess reagent) left over, we will first find moles O2 used up.
moles O2 used = 0.0875 mol propanol x 9 mol O2/2 mol propanol = 0.394 moles O2 used
moles O2 left over = 0.997 mol - 0.394 mol = 0.603 mol O2 left
mass O2 left = 0.603 mol O2 x 32 g/mol = 19.3 g O2 left over
Answer:
1.21 g of Tris
Explanation:
Our solution if made of a solute named Tris
Molecular weight of Tris is 121 g/mol
[Tris] = 100 mM
This is the concentration of solution:
(100 mmoles of Tris in 1 mL of solution) . 1000
Notice that mM = M . 1000 We convert from mM to M
100 mM . 1 M / 1000 mM = 0.1 M
M = molarity (moles of solute in 1 L of solution, or mmoles of solute in 1 mL of solution). Let's determine the mmoles of Tris
0.1 M = mmoles of Tris / 100 mL
mmoles of Tris = 100 mL . 0.1 M → 10 mmoles
We convert mmoles to moles → 10 mmol . 1mol / 1000mmoles = 0.010 mol
And now we determine the mass of solute, by molecular weight
0.010 mol . 121 g /mol = 1.21 g
Water and orange juice
Or orange juice and sodium chloride
With every electron stationed in its own orbital or paired off with each other in the higher energy level, the energy level is balanced and stable. The atoms that utilize this exception are Molybdenum, Chromium, Gold, Silver, and Copper.
Answer: There are 7.4 moles of helium gas present in a 1.85 liter container at the same temperature and pressure.
Explanation:
Given: 
= 2.25 L, 
= 9.0 mol
= 1.85 L, 
= ?
Formula used to calculate the moles of helium are as follows.
Substitute the values into above formula as follows.
Thus, we can conclude that there are 7.4 moles of helium gas present in a 1.85 liter container at the same temperature and pressure.
