To calculate the concentration of silver ions in the simulation, you need to convert the number of dissolved ions into moles. then divide this number of moles by the volume in liters.
how would the concentration of silver ions compare in a 1.0 x 10-16 l saturated solution to a 1.5 x 10-16 l saturated solution? write a hypothesis first, and then test your hypothesis using the simulation. to simulate a saturated solution, put in enough silver bromide that you can clearly see that it stops dissolving. calculate the concentrations of silver ions at both of these volumes.

first one is the correct answer

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g

answer: a tank of height h is filled with water and sits on the ground. water squirts from a hole at a height y above the ground and has a range r. for two y values, 0 and h, r is zero. determine the value of y for which the range will be a maximum.

explanation: