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Answer:
The percent by mass of chromium(II) acetate in the solution is 4.42%.
Explanation:
Molality of the chromium(II) acetate solution = 0.260 m = 0.260 mol/kg
This means that in 1 kg of solution 0.260 moles of chromium(II) acetate are present.
1000 g = 1 kg
So, in 1000 grams of solution 0.260 moles of chromium(II) acetate are present.
Then in 100 grams of solution =
Mass of 0.0260 moles chromium(II) acetate:
= 0.0260 mol × 170 g/mol = 4.42 g
The percent by mass of chromium(II) acetate in the solution is 4.42%.
These 2 isotopes are all of bromine that is naturally found (their percentages add up to 100%). In order to calculate the mean atomic mass, we can use the following calculation: 
. This yields an average atomic mass of 79.903 amu. This procedure is general; whenever we have many types F of a specific genre G, the mean of the genre G can be calculated by:
where
is the proportion of type i in the genre G. In this example, Br79 and Br81 are the 2 types and the genre is Bromine.
where
2.1618 grams of P4 is required to produce 4.21x10^22 molecules of phosphorus trifluoride.
Explanation:
The balanced chemical reaction for formation of PF3 from yellow phosphorus is given by:
P4 + 6F2 ⇒ 4PF3
1 mole of P4 reacts to give 4 moles of PF3
It is given that 4.21x10^22 molecules of PF3 are produced
so the number of moles can be calculated by the relation
number of molecules = number of moles × Avagadro number
number of moles =
n =
n= 0.06989 moles of PF3 is formed.
Applying stoichiometry,
1 mole of P4 gives 4 moles PF3
x mole will produce 0.06989 moles of PF3
=
4x= 0.0698 × 1
x =
x= 0.01745 moles of P4 will be required.
The weight of the phosphorus can be obtained by number of moles × atomic mass of one mole of P4
= 0.01745 × 123.89
= 2.1618 grams of P4.