Answer:
16.89g of PbBr2
Explanation:
First, let us calculate the number of mole of Pb(NO3)2. This is illustrated below:
Molarity of Pb(NO3)2 = 0.595M
Volume = 77mL = 77/1000 = 0.077L
Mole =?
Molarity = mole/Volume
Mole = Molarity x Volume
Mole of Pb(NO3)2 = 0.595x0.077
Mole of Pb(NO3)2 = 0.046mol
Convert 0.046mol of Pb(NO3)2 to grams as shown below:
Molar Mass of Pb(NO3)2 =
207 + 2[ 14 + (16x3)]
= 207 + 2[14 + 48]
= 207 + 2[62] = 207 +124 = 331g/mol
Mass of Pb(NO3)2 = number of mole x molar Mass = 0.046 x 331 = 15.23g
Molar Mass of PbBr2 = 207 + (2x80) = 207 + 160 = 367g/mol
Equation for the reaction is given below:
Pb(NO3)2 + CuBr2 —> PbBr2 + Cu(NO3)2
From the equation above,
331g of Pb(NO3)2 precipitated 367g of PbBr2
Therefore, 15.23g of Pb(NO3)2 will precipitate = (15.23x367)/331 = 16.89g of PbBr2
Answer:
I think its b
Explanation:
but I wouldn't depend on this answer
Assuming that the reactants are:
(NH4)2SO4 (aq) + Ba(NO3)2 (aq)
and the products are:
BaSO4 (s) + 2NH4NO3 (aq),
then you will have to determine which product is insoluble. You should have access to solubility rules to help you determine this.
According to the solubility rules, the following elements are considered insoluble when paired with SO4:
Sr^2+, Ba^2+, Pb^2+, Ag^2+, and Ca^2+
Therefore, the precipitate will be BaSO4 (s).
Answer:
B. Conclusions are based on observations
Explanation:
Fix ur transition, it sounds choppy
