Answer:
Mass = 17.28 g
Explanation:
Given data:
Mass of propane = 35 g
Volume of oxygen = 27 L
Mass of water produced = ?
Solution:
Chemical equation:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
First of all we will calculate the number of moles of oxygen at STP.
PV = nRT
n = PV/RT
n = 1 atm×27 L / 0.0821 atm. L/ mol.K × 273 K
n = 27 atm . L / 22.413 atm. L/mol
n = 1.2 mol
Moles of propane:
Number of moles = mass/ molar mass
Number of moles = 35 g/ 44.1 g/mol
Number of moles = 0.8 mol
Now we will compare the moles of water with propane and oxygen.
C₃H₈ : H₂O
1 : 4
0.8 : 4×0.8 = 3.2 mol
O₂ : H₂O
5 : 4
1.2 : 4/5×1.2 = 0.96 mol
Number of moles of water produced by oxygen are less so it will limiting reactant.
Mass of water:
Mass = number of moles × molar mass
Mass = 0.96 mol × 18 g/mol
Mass = 17.28 g
Answer:
102kPa
Explanation:
760mmHg = 101.325 kPa
12mmHg = x
Upon converting from mmHg to kPa we have;
x = ( 12 mm Hg x 101.325 kPa) / 760mmHg
x = 1.599868421 kPa
Total pressure = 100.8 kPa + 1.599868421 kPa
Total pressure = 102kPa
An inorganic compound is any compound that lacks a carbon atom
Answer:
7836.6 J
Explanation:
The following data were obtained from the question:
Heat of fusion (Hf) = 0.148 KJ/g
Mass (m) = 52.95 g
Heat (Q) required =..?
The heat (Q) required to melt the sample of nephthalene can be obtained as follow:
Q = m·Hf
Q = 52.95 × 0.148
Q = 7.8366 KJ
Finally, we shall convert 7.8366 KJ to Joule (J) in order to obtain the desired result. This can be obtained as follow:
1 kJ = 1000 J
Therefore,
7.8366 KJ = 7.8366 KJ × 1000 J / 1 KJ
7.8366 KJ = 7836.6 J
Therefore, 7.8366 KJ is equivalent to 7836.6 J
Thus, 7836.6 J of heat energy is required to melt the sample of nephthalene.
Electrons<span> are NOT in circular orbits around nucleus. -The</span>atomic<span> orbital </span>describes<span> the probable location of the electrion </span>Quantum Mechanical Model of the Atom<span> Page 3 There are different kinds of </span>atomic orbitals<span> that differ in the amount of energy and shapes (where the </span>electron<span> probably is).
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