Answer:
Ammonium bromide can be prepared by the direct action of hydrogen bromide on ammonia. It can also be prepared by the reaction of ammonia with iron(II) bromide or iron(III) bromide, which may be obtained by passing aqueous bromine solution over iron filings.
Explanation:
please mark me as brainliest thank you 
 
        
             
        
        
        
<span>0 °C. 8 °C. 29 °C. 15 °C.</span>
        
             
        
        
        
Answer:
28.75211 kj
Explanation:
Given data:
Mass of iron bar = 841 g
Initial temperature = 84°C
Final temperature = 7°C
Heat released = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
specific heat capacity of iron is 0.444 j/g.°C
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 7°C - 84°C 
ΔT = -77°C
By putting values,
Q = 841 g × 0.444 j/g.°C × -77°C
Q = 28752.11 j
In Kj:
28752.11 j × 1 kJ / 1000 J
28.75211 kj
 
        
             
        
        
        
Answer:
11.9 is the pOH of a 0.150 M solution of potassium nitrite.
Explanation:
Solution :  Given,
Concentration (c) = 0.150 M
Acid dissociation constant = 
The equilibrium reaction for dissociation of 
 (weak acid) is,
                            
initially conc.         c                       0         0
At eqm.              
                
        
First we have to calculate the concentration of value of dissociation constant 
.
Formula used :

Now put all the given values in this formula ,we get the value of dissociation constant 
.



By solving the terms, we get

No we have to calculate the concentration of hydronium ion or hydrogen ion.
![[H^+]=c\alpha=0.150\times 0.0533=0.007995 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dc%5Calpha%3D0.150%5Ctimes%200.0533%3D0.007995%20M)
Now we have to calculate the pH.
![pH=-\log [H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH%5E%2B%5D)


pH + pOH = 14
pOH =14 -2.1 = 11.9
Therefore, the pOH of the solution is 11.9