Step one write the equation for dissociation of AgNO3 and NaCl
that is AgNO3-------> Ag+ + NO3-
NaCl--------> Na+ + Cl-
then find the number of moles of each compound
that is for AgNO3 = ( 1.4 x10^-3 ) x 25/1000= 3.5 x10^-5 moles
Nacl= (7.5 x10^-4)x 60/1000= 4.5 x10^-5 moles
from mole ratio the moles of Ag+= 3.5 x10^-5 moles and that of Cl-= 4.5 x10^-4 moles
then find the total volume of the mixture
that is 25ml + 60 Ml =85ml = 0.085 liters
The Ksp of Agcl = (Ag+) (cl-), let the concentration of Ag+ be represented by x and also the concentration be represented by x
ksp (1.8 x10^-10) is therefore= x^2
find the square root x=1.342 x10^-5
Ag+ in final mixture is = moles of Ag+/total volume - x
that is {(3.5 x10^-5)/0.085} - 1.342 x10^-5=3.98x10^-4
Cl- in the final mixture is =(4.5 x10^-5 /0.085) - 1.342 x10^-5= 5.16 x10^-4
Your answer is C forms of the same element that differ in the number of neutrons in each atom because if the number of neutrons change the whole element changes.
<span>First - you need the empirical formula.
So, assume you have 100 g of the compound.
If so, you'll have 54.53 gram of C, 9.15 g of H and 36.32 g of O. Find the number of moles of each.
54.53 g C (1 mole C / 12.01 g C) = 4.540
9.15 g H (1 mole H / 1.008 g H) = 9.077
36.32 g O (1 mole O / 15.9994 g O) = 2.270
Take the smallest number found and divide the others by it to get the empirical formula.
4.540/2.270 = 2.
9.077/2.270 = 4.
2.270/2.270 =1.
So, that gives you the empirical formula of C2H4O.
Find the weight of this compound. C = 12, H = 1, O = 16. So, C2H4O is 44 amu.
132/44 = 3.
So, 3 (C2 H4 O) = C6H12O3 = molecular formula.</span>
Answer:
Negative
Explanation:
Any atom gaining electrons will get a negative change since the charge of the electron is negative.
Among the given choices NH₃/NH₄Cl is the best choice to prepare pH = 9.0 because pKa of the salt of NH₄<span>Cl nearly equals to 9 (pKb = 4.75, hence pKa = 14 - pKb = 14 - 4.75 = 9.25). A buffer can't change its pH +/- pKa, so this combination is best to prepare desired pH.</span>
