Answer:
A. N₂(g) + 3H₂(g) -----> 2NH₃    exothermic
B. S(g) + O₂(g) --------> SO₂(g)    exothermic
C. 2H₂O(g) --------> 2H₂(g) + O₂(g) endothermic
D. 2F(g) ---------> F₂(g) exothermic
Explanation:
The question says predict not calculate. So you have to use your chemistry knowledge, experience and intuition.
A. N₂(g) + 3H₂(g) -----> 2NH₃    is exothermic because the Haber process gives out energy 
B. S(g) + O₂(g) --------> SO₂(g)    is exothermic because it is a combustion. The majority, if not all, combustion give out energy.
C. 2H₂O(g) --------> 2H₂(g) + O₂(g) is endothermic because it is the reverse reaction of the combustion of hydrogen. If the reverse reaction is exothermic then the forward reaction is endothermic  
D. 2F(g) ---------> F₂(g) is exothermic because the backward reaction is endothermic. Atomisation is always an endothermic reaction so the forward reaction is exothermic
 
        
             
        
        
        
Answer:  0.745 g of  will be produced from  1.08 g of sodium sulfate
 will be produced from  1.08 g of sodium sulfate
Explanation:
To calculate the moles :
 
    
 
  
 
 
 is the limiting reagent as it limits the formation of product and
 is the limiting reagent as it limits the formation of product and  is the excess reagent.
 is the excess reagent.
According to stoichiometry :
3 moles of  produce = 3 moles of
 produce = 3 moles of  
Thus 0.0076 moles of  will require=
 will require= of
  of  
Mass of  
Thus 0.745 g of  will be produced from  1.08 g of sodium sulfate
 will be produced from  1.08 g of sodium sulfate
 
        
             
        
        
        
Molecular mass= (14.01∗1)+(1.008∗3)
                         14.01+3.024=17.03g/mol
        
             
        
        
        
Because during combustion reaction, heat energy is released and it's this energy that is converted to work
 
        
                    
             
        
        
        
Answer:
3
Explanation:
Applying,
 = R/R'............... Equation 1
 = R/R'............... Equation 1
Where n' = number of halflives that have passed, R = Original atom of the substance, R' = atom of the substance left after decay.
From the question,
Given: R = 40 atoms, R' = 5 atoms
Substitute these values into equation 1
 = 40/5
 = 40/5
 = 8
 = 8
 = 2³
 = 2³
Equation the base,
n' = 3