Answer:
0.51M
Explanation:
Given parameters:
Initial volume of NaBr = 340mL
Initial molarity = 1.5M
Final volume = 1000mL
Unknown:
Final molarity = ?
Solution;
This is a dilution problem whereas the concentration of a compound changes from one to another.
In this kind of problem, we must establish that the number of moles still remains the same.
number of moles initially before diluting = number of moles after dilution
Number of moles = Molarity x volume
Let us find the number of moles;
Number of moles = initial volume x initial molarity
Convert mL to dm³;
1000mL = 1dm³
340mL gives = 0.34dm³
Number of moles = initial volume x initial molarity = 0.34 x 1.5 = 0.51moles
Now to find the new molarity/concentration;
Final molarity = = = 0.51M
We can see a massive drop in molarity this is due to dilution of the initial concentration.
We have to calculate the percentage (by mass) of Ba in the mixture of BaBr₂ and inert material.
Given, bromide ion is precipitated as AgBr by the addition of an excess of silver nitrate as per following reaction:
BaBr₂(aq) + 2AgNO₃(aq)→2AgBr(s)+Ba(NO₃)₂(aq). From this balanced chemical reaction, it is clear that one mole BaBr₂ reacts with two moles of AgNO₃ and gives two moles of AgBr. We have to calculate 0.6226 g AgBr contains how many moles of AgBr. Using that information we can get how many moles of BaBr₂ reacted to give 0.6226g AgBr and one mole BaBr₂ contains one mole of Ba. By multiplying number of moles of Ba with atomic mass of Ba we can get amount of Ba present in the mixture. Accordingly we can calculate mass percentage of Ba in the mixture.
Atomic mass of Ba= 137.327 g.mol, Molecular mass of BaBr₂=297.1 g/mol and molecular mass of AgBr=187.7 g/mol. Mass of AgBr is 0.6226 g which contains 0.6226/187.7 mole= 3.31X10⁻³ moles of AgBr. So, moles of BaBr₂ reacts= (3.31X10⁻³)/2 moles= 1.65X10⁻³ moles. One mole BaBr₂ contains one mole of Ba. So, 1.65X10⁻³ moles of BaBr₂ contains 1.65X10⁻³ moles of Ba atoms whose mass is=1.65X10⁻³X137.327g=0.2265 g. Out of 0.7207 g sample amount of Ba present is 0.2265 g. So, mass percentage is =31.42 %
.
CBr4 the carbon would be taken electro positive as plus4 and bromine -1 as its in group 7
Answer:
Cu + Fe 3 Pb 2 +
Explanation:
the most reactive metal is the strongest reducing agent but weakest oxidizing agent. And therefore copper being the least reactive turns to be the strongest oxidizing agent followed by iron then lead.