What noble gas would be part of
configuation notion for mn

Argon

Explanation:

Manganese is an atom with 25 electrons in its orbitals. The orbital notation is shown below:

                        1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s²

The closest noble gas to manganese is argon with a configuration of:

                         1s² 2s² 2p⁶ 3s² 3p⁶

Therefore, based on the noble gase Ar, we can write Mn as:

                                  [Ar]3d⁵ 4s²

We can solve this problem by assuming that the decay of cyclopropane follows a 1st order rate of reaction. So that the equation for decay follows the expression:

A = Ao e^(- k t) 

Where,

A = amount remaining at time t = unknown (what to solve for) 
Ao = amount at time zero = 0.00560 M 
k = rate constant
t = time = 1.50 hours or 5400 s 

The rate constant should be given in the problem which I think you forgot to include. For the sake of calculation, I will assume a rate constant which I found in other sources:

k = 5.29× 10^–4 s–1                     (plug in the correct k value)

Plugging in the values in the 1st equation:

A = 0.00560 M * e^(-5.29 × 10^–4 s–1 * 5400 s )

A = 3.218 × 10^–4 M           (simplify as necessary)