1) Balanced chemical reaction:
2H2 + O2 -> 2H20
Sotoichiometry: 2 moles H2: 1 mol O2 : 2 moles H2O
2) Reactant quantities converted to moles
H2: 5.00 g / 2 g/mol = 2.5 mol
O2: 50.0 g / 32 g/mol = 1.5625 mol
Limitant reactant: H2 (because as per the stoichiometry it will be consumed with 1.25 mol of O2).
3) Products
H2 totally consumed -> 0 mol at the end
O2 = 1.25 mol consumed -> 1.5625 mol - 1.25 mol = 0.3125 mol at the end
H2O: 2.5 mol H2 produces 2.5 mol H2O -> 2.5 mol at the end.
Total number of moles: 0.3125mol + 2.5 mol = 2.8125 mol
4) Pressure
Use pV = nRT
n = 2.8125
V= 9 liters
R = 0.082 atm*lit/K*mol
T = 35 C + 273.15 = 308.15K
p = nRT/V = 7.9 atm
Rutherford, Geiger and Marsden's experiment proved that every atom has a nucleus and that this nucleus is of positive charge and contains the most of the mass of the atom. 0.005% of the volume occupied by the electrons is the volume of the nucleus.
Answer:
1.43 M
Explanation:
We'll begin by calculating the number of mole of the solid. This can be obtained as follow:
Mass of solid = 8.60 g
Molar mass of solid = 21.50 g/mol
Mole of solid =?
Mole = mass / molar mass
Mole of solid = 8.60 / 21.50
Mole of solid = 0.4 mole
Next, we shall convert 280 mL to litre (L). This can be obtained as follow:
1000 mL = 1 L
Therefore,
280 mL = 280 mL × 1 L / 1000 mL
280 mL = 0.28 L
Thus, 280 mL is equivalent to 0.28 L.
Finally, we shall determine the molarity of the solution. This can be obtained as illustrated below:
Mole of solid = 0.4 mole
Volume = 0.28 L
Molarity =?
Molarity = mole / Volume
Molarity = 0.4 / 0.28
Molarity = 1.43 M
Thus, the molarity of the solution is 1.43 M.
Answer: the bonds in the methane and oxygen come apart, the atoms rearrange and then re-bond to form water and carbon dioxide
Explanation:^
Answer:
Explanation:
Hello there!
In this case, given the neutralization of the acetic acid as a weak one with sodium hydroxide as a strong base, we can see how the moles of the both of them are the same at the equivalence point; thus, it is possible to write:
Thus, we solve for the molarity of the acid to obtain:
Regards!
