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Trophic level II Members
Answer:
t = 5.7634 s
Explanation:
- A → Pdts
- - rA = K (CA)∧α = - δCA/δt
∴ T = 400°C
∴ α = 1 ....first-order
∴ CAo = 0.950 M
∴ CA = 0.300 M
⇒ t = ?
⇒ - δCA/δt = K*CA
⇒ - ∫δCA/CA = K*∫δt
⇒ Ln (CAo/CA) = K*t
⇒ t = Ln(CAo/CA) / K
⇒ t = (Ln(0.950/0.300)) / (0.200 s-1)
⇒ t = 1.1527 / 0.200 s-1
⇒ t = 5.7634 s
Answer:

Explanation:
Hello!
In this case for the solution you are given, we first use the mass to compute the moles of CuNO3:

Next, knowing that the molarity has units of moles over liters, we can solve for volume as follows:

By plugging in the moles and molarity, we obtain:

Which in mL is:

Best regards!