Answer:
The answer to your question is 0.4 moles of Oxygen
Explanation:
Data
Octane (C₈H₈)
Oxygen (O₂)
Carbon dioxide (CO₂)
Water (H₂O)
moles of water = ?
moles of Oxygen = 1
Balanced chemical reaction
C₈H₈ +10O₂ ⇒ 8CO₂ + 4H₂O
Reactant Element Products
8 C 8
8 H 8
20 O 20
Use proportions to solve this problem
10 moles of Oxygen ----------------- 4 moles of water
1 mol of Oxygen ------------------ x
x = (4 x 1) / 10
x = 4 / 10
x = 0.4 moles of water
Answer:a quantum absorption of energy
Explanation:
Bohr’s model explains the spectral lines .While the electron of the atom remains in the ground state, its energy is unchanged. When the atom absorbs one or more quanta of energy, the electron moves from the ground state orbit to an excited state and when the atom relaxes back to a lower energy state, it releases energy that is again equal to the difference in energy of the two orbits.
Answer:
This is very easy Cuz We have 2Na2O We have O2 so thats molecule of Oxygen and its same on product We need to balance Na on start We have 1 on product We have 2 so Just put 2 at start....
Explanation:
Sorry for bad english not my first language :(
2Na+O2-->2Na20
B its viscosity decreases
Answer:
pH = 4.543
Explanation:
- CH3CH2COOH + H2O ↔ CH3CH2COO- + H3O+
- pKa = - Log Ka
∴ Ka = [H3O+][CH3CH2COO-]/[CH3CH2COOH]
∴ pKa = 4.87
⇒ Ka = 1.349 E-5 = [H3O+][CH3CH2COO-]/[CH3CH2COOH]
added 300 mL 0f 0.02 M NaOH:
⇒ <em>C</em> CH3CH2COOH = ((0.200 L)(0.15 M)) - ((0.300 L)(0.02 M))/(0.3 + 0.2)
⇒ <em>C</em> CH3CH2COOH = 0.048 M
⇒ <em>C</em> NaOH = (0.300 L)(0.02 M) / (0.3 +0.2) = 0.012 M
mass balance:
⇒ 0.048 + 0.012 = 0.06 M = [CH3CH2COO-] + [CH3CH2COOH].......(1)
charge balance:
⇒ [H3O+] + [Na+] = [CH3CH2COO-]
∴ [Na+] = 0.02 M
⇒ [CH3CH2COO-] = [H3O+] + 0.02 M.............(2)
(2) in (1):
⇒ [CH3CH2COOH] = 0.06 M - 0.02 M - [H3O+] = 0.04 M - [H3O+]
replacing in Ka:
⇒ 1.349 E-5 = [H3O+][([H3O+] + 0.02) / (0.04 - [H3O+])
⇒ (1.349 E-5)(0.04 - [H3O+]) = [H3O+]² + 0.02[H3O+]
⇒ 5.396 E-7 - 1.349 E-5[H3O+] = [H3O+]² + 0.02[H3O+]
⇒ [H3O+]² + 0.02001[H3O+] - 5.396 E-7 = 0
⇒ [H3O+ ] = 2.867 E-5 M
∴ pH = - Log [H3O+]
⇒ pH = 4.543
