Answer:
"0.60 g" is the appropriate solution.
Explanation:
The given values are:
Volume of base,
= 30 ml
Molarity of base,
= 0.05 m
Molar mass of acid,
= 400 g/mol
As we know,
⇒ 
On substituting the values, we get
⇒ 
⇒ 
⇒
hence,
⇒ 
On substituting the values, we get
⇒ 
⇒ 
⇒ 
Answer:
the catalyst is the two gray dots
Answer:
NH4Br + AgNO3 —> AgBr + NH4NO3
Explanation:
When ammonium bromide and silver(I) nitrate react, the following are obtained as shown below:
NH4Br(aq) + AgNO3(aq) —>
In solution, NH4Br(aq) and AgNO3(aq) will dissociate as follow:
NH4Br(aq) —> NH4+(aq) + Br-(aq)
AgNO3(aq) —> Ag+(aq) + NO3-(aq)
The double displacement reaction will occur as follow:
NH4+(aq) + Br-(aq) + Ag+(aq) + NO3-(aq) —> Ag+(aq) + Br-(aq) + NH4+(aq) + NO3-(aq)
NH4Br(aq) + AgNO3(aq) —> AgBr(s) + NH4NO3(aq)
Count up the number of electrons in each orbital. theres ten electrons which means it has 10 protons. the element would be neon