Explanation:
According to law of conservation of mass, mass of the reactants is equal to the mass of products in a chemical equation. As mass can neither be created nor it can be destroyed but it can be transformed from one form to another.
As it is given that hydrogen and in excess oxygen is reacting that leads to the formation of water. Hence, the chemical reaction equation will be as follows.

Since, it is given that 4 mol of hydrogen is reacting with excess of oxygen and gives 2 moles of water.
Hence, number of moles of water produced is calculated as follows.
4 mol of 
= 4 moles of 
Thus, we can conclude that 4 moles of water you can produce from 4.0 mol of hydrogen and excess oxygen.
The random molecular movement from higher concentration to lower concentration
Answer:
156 Hydrogen atoms
Explanation:
<u>Any acyclic alkane has a molecular formula that can be expressed as</u>:
CₙH₂ₙ₊₂
Where <em>n</em> is any integer and the number of carbon atoms. For example, Propane has 3 carbon atoms, this means it would have [2*3+2] 8 hydrogen atoms, resulting with a formula of C₃H₈.
An acyclic alkane with 77 carbon atoms would thus have:
2*77 + 2 = 156 hydrogen atoms
The rate of a chemical reaction can be raised by increasing the surface area of a solid reactant. This is done by cutting the substance into small pieces, or by grinding it into a powder.
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, 