Achemist prepares a solution of nickel(ii) chloride (nicly) by measuring out 61.3 umol of nickel(ii) chloride into a 200 ml volumetric flask and filling the flask to the mark with water. calculate the concentration in umol/l of the chemist's nickel(ii) chloride solution. round your answer to 3 significant digits. x 5 ? explanation check 2009 mew econ aires reserved. terms of use

The concentration of nickel(II) chloride in μmol/l will be 307 μmol/l

Explanation:

If in 200 ml you have 61.3 μmol, then, in 1000 ml ( 1 l) you will have:

1000 ml * (61.3 μmol / 200 ml) = 306.5 μmol.

Rounded to 3 significant digits: 307 μmol.

The concentration will be 307 μmol/l or 307 μM.

it is to be noted that an alpha particle is an helium nucleus. so it contains 2 protons 2 neutrons and has a mass number of 4. when alpha decay particles are ejected from the nucleus it gets transformed into an atom which has atomic number less than 2 and a mass number less than 4.

an equation that could represent alpha decay is radium 226 decay of alpha particle forms radon 222.

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explanation:

i think option no d is the correct answer