The awnser would be c i got you fam
Answer:
M of HI = 5.4 M.
Explanation:
- We have the rule: at neutralization, the no. of millimoles of acid is equal to the no. of millimoles of the base.
<em>(XMV) acid = (XMV) base.</em>
where, X is the no. of (H) or (OH) reproducible in acid or base, respectively.
M is the molarity of the acid or base.
V is the volume of the acid or base.
<em>(XMV) HI = (XMV) Ca(OH)₂.</em>
For HI; X = 1, M = ??? M, V = 25.0 mL.
For Ca(OH)₂, X = 2, M = 1.5 M, V = 45.0 mL.
<em>∴ M of HI = (XMV) Ca(OH)₂ / (XV) HI</em> = (2)(1.5 M)(45.0 mL) / (1)(25.0 mL) = <em>5.4 M.</em>
The mass of 8 mol of ¹²C is 96 g.
Mass = 8 mol × (12 g/1 mol) = 96 g
Answer:
Yes, a precipitate of PbS forms
Explanation:
The equation of the reaction is given as:
Pb(NO₃)₂ + Na₂S → PbS + 2NaNO₃
The lead sulfide forms a precipitate in the aqeous solution.
Precipitation is a form of reaction in which ions combines to form a solid precipiate. Most double displacement reactions in which ionic compounds are the reactants results in formation of a precipitate as the product.
There are rules of solubility which guides a reaction that would lead to the formation of a precipitate. The mos applicable of the rules to the reaction stated above is that "carbonates, phosphates, sulfides, oxides and hydroxides are insolube". The sulfide of lead formed in the product is therefore insoluble.
