Answer:
Explanation:
From the given information:
Camphor may be reduced as readily in the presence of sodium borohydride(NaHB4). The resulting compound which is stereoselective requires 1 mole of sodium borohydride (NaHB4) to reduce 1 mole of camphor in this reaction. The reaction is shown below.
Through the reduction process of camphor, the reducing agent can reach the carbonyl face with a one-carbon linkage. The product stereoisomer is known as borneol.
If the molecular weight of camphor = 152.24 g/mol
and it mass = 200 mg
The its no of moles = 200 mg/ 152.24 g/mol
= 1.3137 mmol
Now the amount of the required mmol for NaBH4 to be consumed in the reaction = 5.2 × 1.3137 mmol
= 6.831 mmol
since the molar mass of NaBH4 = 37.83 g/mol
Then, using the same formula:
No of moles = mass/molar mass
mass = No of moles × molar mass
mass = 6.831 mmol × 37.83 g/mol
mass of NaBH4 used = 258.42 mg
Answer : The concentration after 17.0 minutes will be, 
Explanation :
The expression for first order reaction is:
![[C_t]=[C_o]e^{-kt}](https://tex.z-dn.net/?f=%5BC_t%5D%3D%5BC_o%5De%5E%7B-kt%7D)
where,
= concentration at time 't' (final) = ?
= concentration at time '0' (initial) = 0.100 M
k = rate constant = 
t = time = 17.0 min = 1020 s (1 min = 60 s)
Now put all the given values in the above expression, we get:
![[C_t]=(0.100)\times e^{-(5.40\times 10^{-3})\times (1020)}](https://tex.z-dn.net/?f=%5BC_t%5D%3D%280.100%29%5Ctimes%20e%5E%7B-%285.40%5Ctimes%2010%5E%7B-3%7D%29%5Ctimes%20%281020%29%7D)
![[C_t]=4.05\times 10^{-4}M](https://tex.z-dn.net/?f=%5BC_t%5D%3D4.05%5Ctimes%2010%5E%7B-4%7DM)
Thus, the concentration after 17.0 minutes will be, 
1. 1 M , 2 M , 1 M
2. 10 mol , 0.1 mol , 0.5 mol
3. 0.5 L , 6.6 L , 5/21 L
M = mol/L