Need help please: How many mL of 0.150 Molarity CaBr2 solution must be used to react with 50 mL of 0.115 Molarity AgNO3? CaBr2 + 2AgNO3 --> Ca(No3)2 + 2AgBr

0.019 L or 19.1 mL

Explanation:

Given the equation of the reaction; we have:

CaBr2 + 2AgNO3 --> Ca(No3)2 + 2AgBr

We must first obtain the number of moles of AgNO3 reacted. This can be obtained from;

Concentration of AgNO3 reacted= 0.115 M

Volume of AgNO3 reacted= 50 ml

Since

Number of moles of AgNO3= concentration of AgNO3 × volume of AgNO3

Number of moles = 0.115 M × 50/1000 = 5.75 ×10^-3 moles

From the reaction equation;

2 moles of AgNO3 reacts with 1 mole of CaBr2

5.75 ×10^-3 moles of AgNO3 reacts with 5.75 ×10^-3 moles × 1 /2 = 2.875×10^-3 moles of CaBr2

Now;

Number of moles of CaBr2= concentration of CaBr2 × volume of CaBr2

Number of moles of CaBr2= 2.875×10^-3 moles

Volume of CaBr2=

Concentration of CaBr2= 0.150 M

Hence;

Volume of CaBr2= number of moles of CaBr2/ concentration of CaBr2

Volume of CaBr2= 2.875×10^-3 moles / 0.150

Volume of CaBr2= 0.019 L or 19.1 mL

the correct answer would be b. amelia, because the particals of gas move really fast (charged particals) and gas also has a lot of room (particals move independently) so the only answer that has both of those is amelia's.