Answer:
B: increase.
Explanation:
When we are considering two gases A and B in a container at room temperature .
We have to find the change on  rate of reaction when the number of molecules of gases A is doubled
Let [A]=a and [B]=b
A+B
 product
Rate of reaction
![R_1=k[A][B]=kab](https://tex.z-dn.net/?f=R_1%3Dk%5BA%5D%5BB%5D%3Dkab)
We know that concentration is increases with increase in number of moles 
When the number of molecules of gases A is doubled then concentration of gases A increases.
Therefore ,[A]=2a
Rate of reaction 


Hence, the rate of reaction is  2 times the initial rate of reaction.Therefore, the rate of reaction will increase when the number of molecules of gases A is doubled.
Answer: B: increase.
 
        
             
        
        
        
To be honest I don’t even know
        
                    
             
        
        
        
Answer:
4.03dm³
Explanation:
The reaction expression is given as: 
        3H₂   +   N₂   →    2NH₃  
   Volume of hydrogen  = 12dm³  
AT rtp: 
              1 mole of gas occupies volume of 22.4dm³  
              x mole of hydrogen will occupy a volume of 12dm³
      Number of moles of hydrogen  = 
   = 0.54mole 
From the balanced reaction equation: 
             3 mole of hydrogen gas combines with 1 mole of Nitrogen gas
          0.54 mole of hydrogen as will therefore combine with 
   = 0.18moles of nitrogen gas 
 Since ; 
                      1 mole of gas occupies a volume of 22.4dm³
                0.18moles of Nitrogen gas will occupy 0.18 x 22.4  = 4.03dm³