In order to balance an equation, we apply the principle of conservation of mass, which states that mass can neither be created nor destroyed. Therefore, the mass of an element before and after a reaction remains constant. Here, the balanced equation becomes:
4Al + 3O₂ → 2Al₂O₃
The coefficients are 4, 3 and 2.
Answer:
At a front, the two air masses have different densities, based on temperature, and do not easily mix. One air mass is lifted above the other, creating a low pressure zone.
Explanation:
Hope this helps!
Answer:
Mass = 179.9 g
Explanation:
Given data:
Volume of solution = 450 mL
Molarity of solution = 2.00 M
Mass in gram required = ?
Solution:
Volume of solution = 450 mL× 1 L / 1000 mL = 0.45 L
Molarity = number of moles of solute/ Volume of solution in L
2.00 M = number of moles of solute / 0.45 L
Number of moles of solute = 2.00 M × 0.45 L
M = mol/L
number of moles of solute = 0.9 mol
Mass of CaBr₂ in gram:
Mass = number of moles × molar mass
Mass = 0.9 mol ×199.89 g/mol
Mass = 179.9 g
Answer:
What will determine the number of moles of hydronium in an aqueous solution of a strong monoprotic acid? The amount of acid that was added.
Explanation:
Answer:
Kc = Kc = 8.0 * 10^9
Kp = 5.5 *10^5
Explanation:
Step 1: Data given
Temperature = 25.0 °C
Number of moles Fe = 1.0 moles
Number of moles O2 = 1.0 * 10^-3 moles
Number of moles Fe2O3 = 2.0 moles
Volume = 2.0 L
Step 2: The balanced equation
4Fe(s) + 3O2(g) ⇌ 2Fe2O3(s)
Step 3: Calculate molarity
Molarity = moles / volume
[Fe] = 1.0 moles / 2.0 L
[Fe] = 0.5 M
[O2] = 0.001 moles / 2.0 L
[O2] = 0.0005 M
[Fe2O3] = 2.0 moles / 2.0 L
[Fe2O3] = 1.0 M
Step 4: Calculate Kc
Kc =1/ [O2]³
Kc = 1/0,.000000000125
Kc = 8.0 * 10^9
Step 5: Calculate Kp
Kp = Kc*(R*T)^Δn
⇒with Kc = 8.0*10^9
⇒with R = 0.08206 L*atm /mol*K
⇒with T = 298 K
⇒with Δn = -3
Kp = 8.10^9 *(0.08206 * 298)^-3
Kp = 5.5 *10^5
