A0.3146-g sample of a mixture of nacl(s) and kbr(s) was dissolved in water. the resulting solution required 45.30 ml of 0.08765 m agno3(aq) to precipitate the cl–(aq) and br–(aq) as agcl(s) and agbr(s). calculate the mass percentage of nacl(s) in the mixture.

Let mass of NaCl in the mixture = x g

Mass of KBr = 0.3146 -x

Moles of NaCl = Mass / Molar mass = x / 58.5

Mole sof KBr = 0.3146-x / 119

Moles of AgNO3 added = molarity X volume (L) = 0.08765 X 45.30 / 1000 = 0.00397 moles

Out of these say moles of AgNO3 reacted with Cl- = y

So moles of AgNO3 reacted with Br- = 0.00397- y

We know that

x / 58.5  = y  (1)

0.3146-x / 119 = 0.00397 - y (2)

Putting (1) in (2)

0.3146 -x /119 = 0.00397 - x / 58.5

18.40 - 58.5x = 27.64 - 119x

x = 0.153 grams  = Mass of NaCl

MAss % of NaCl  = Mass of NaCl X 100 / total mass

                  = 0.153 X 100 / 0.3146 = 48.63%

it can be observed without changing the identity of the object.

it depends on what substance you're talking about as well as many other factors. in quantum mechanics, the observation of particles affects the observers ability to measure certain properties. for example: you can know the velocity of a particle, but you can't know it's position; or you can know it's position, but not it's velocity. it's like trying to find out where a grain of sand is by throwing other grains of sand at it. once you find out where it is, it's been moved. it's also like trying to find an answer to the question of whether or not a tree falling in a forest makes a sound if nobody is there to hear it. paradoxically, the only way to find out is to be there and hear it.