Answer:
1.25 M
Explanation:
Step 1: Given data
Mass of KI (solute): 20.68 g
Volume of the solution: 100 mL (0.100 L)
Step 2: Calculate the moles of solute
The molar mass of KI is 166.00 g/mol.
20.68 g × 1 mol/166.00 g = 0.1246 mol
Step 3: Calculate the molar concentration of KI
Molarity is equal to the moles of solute divided by the liters of solution.
M = 0.1246 mol/0.100 L= 1.25 M
None of the alpha particles fired at the foil are being repelled back, like they were in the Rutherford atom simulation.I hope this correct.
Answer:
1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹
2. 0.58 mol
Explanation:
1.Given ΔO₂/Δt…
2H₂O₂ ⟶ 2H₂O + O₂
-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt
d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹
d[H₂O]/dt = 2d[O₂]/dt = 2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = 6.6 × 10⁻³mol·L⁻¹s⁻¹
2. Moles of O₂
(a) Initial moles of H₂O₂
(b) Final moles of H₂O₂
The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.
(c) Moles of H₂O₂ reacted
Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol
(d) Moles of O₂ formed
M1 = 17.45 M
M2 = 0.83 M
V2 = 250 ml
M1. V1= M2. V2
V1 = (M2. V2)/M1 = (0.83× 250)/ 17.45= 11.89 ml
Answer: option (1) decreases.
Explanation:
May be you have experienced that: when you go to the beach, where the atmposhpere pressure is greater than the atmosphere pressure in places that are at higher altitudes, the water takes longer to boil. That is because the boiling temperature is greater, and you need more total heat (more time) to permit the liquid to reach that temperature.
The reason why that happens is because substances boil when the vapor pressure (the pressure of the particles of vapor over the liquid) equals the atmosphere pressure. So, when the atmposhere pressure increases, the temperature at which the vapor pressure reaches the atmosphere pressure also increases, and when the atmosphere pressure decreases, the temperature at which the vapor pressure reaches the atmosphere pressure decreases.
