<span>You are given O2 and C3H8, this is a combustion
reaction. The chemical reaction is C3H8 + 10O2 à 3CO2 + 4H2O. You are also given the molar mass
of O2 which is 32.00 g/mol and C3H8 which is 44.1 g/mol. You are required to
find the mass of O2 in grams. Since you have the reaction, oe mole of C3H8 is
required to completely react 10 moles of O2. So,</span>
0.025g C3H8(1 mol C3H8/44.1 g C3H8)(10 mol O2/1
mol C3H8)(32 g O2/1 mol O2) = <u>0.1802 g O2
</u>
<span> </span>
Answer:
Chromatography is a method by which a mixture is separated by distributing its components between two phases. The stationary phase remains fixed in place while the mobile phase carries the components of the mixture through the medium being used.
Have a great day! ^^
A neutron has the same mass as a proton my nilla
Answer:
The Ka is 9.11 *10^-8
Explanation:
<u>Step 1: </u>Data given
Moles of HX = 0.365
Volume of the solution = 835.0 mL = 0.835 L
pH of the solution = 3.70
<u>Step 2:</u> Calculate molarity of HX
Molarity HX = moles HX / volume solution
Molarity HX = 0.365 mol / 0.835 L
Molarity HX = 0.437 M
<u />
<u>Step 3:</u> ICE-chart
[H+] = [H3O+] = 10^-3.70 = 1.995 *10^-4
Initial concentration of HX = 0.437 M
Initial concentration of X- and H3O+ = 0M
Since the mole ratio is 1:1; there will react x M
The concentration at the equilibrium is:
[HX] = (0.437 - x)M
[X-] = x M
[H3O+] = 1.995*10^-4 M
Since 0+x = 1.995*10^-4 ⇒ x=1.995*10^-4
[HX] = 0.437 - 1.995*10^-4 ≈ 0.437 M
[X-] = x = 1.995*10^-4 M
<u>Step 4: </u>Calculate Ka
Ka = [X-]*[H3O+] / [HX]
Ka = ((1.995*10^-4)²)/ 0.437
Ka = 9.11 *10^-8
The Ka is 9.11 *10^-8
Answer:
For carbon-12 is: 6c^12
For carbon-13 is: 6c^13
For carbon-14 is 6c^14
Note: the number 6 is the lower subscript in all three.