A chemist must dilute 47.2 mL of 150. mM aqueous sodium nitrate solution until the concentration falls to . He'll do this by add
1 answer:
Answer:
0.295 L
Explanation:
It seems your question lacks the final concentration value. But an internet search tells me this might be the complete question:
" A chemist must dilute 47.2 mL of 150. mM aqueous sodium nitrate solution until the concentration falls to 24.0 mM. He'll do this by adding distilled water to the solution until it reaches a certain final volume. Calculate this final volume, in liters. Be sure your answer has the correct number of significant digits. "
Keep in mind that if your value is different, the answer will be different as well. However the methodology will remain the same.
To solve this problem we can<u> use the formula</u> C₁V₁=C₂V₂
Where the subscript 1 refers to the concentrated solution and the subscript 2 to the diluted one.
- 47.2 mL * 150 mM = 24.0 mM * V₂
- V₂ = 295 mL
And <u>converting into L </u>becomes:
- 295 mL *
= 0.295 L
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Answer:
a) ²²⁴Ra₈₈ ---> α + ²²⁰Rn₈₆ + Q
b) the Q-value of this reaction is 5.789 MeV
c) the energies (in MeV) of the two associated alpha particles are; 5.69 MeV and 5.449 Mev
Explanation:
a)
The decay equation for the alpha decay is expressed as;
²²⁴Ra₈₈ ---> α + ²²⁰Rn₈₆ + Q
b)
Calculate the Q-value (in MeV) of this reaction.
Q = Mparent - Mdaughter -Mg
Q = MRa - MRn -Mg
= 224.020202 - 220.011384 - 4.00260305
= 0.00621495 amu
= 5.789 MeV
therefore the Q-value of this reaction is 5.789 MeV
c)
Energy of alpha particle is expressed as;
E∝ = MQ / ( m + M)
now this is the maximum energy available for the daughter, ²²⁰Rn going to the ground state;
The energy of the alpha particle gives;
E∝ = 220(5.789) / ( 4 + 220) = 5.69 MeV
as given in the question,The other less frequent alpha occurring 5.5% of the time leaves the daughter nucleus in an excited state of 0.241 MeV above the ground state.
Therefore the energy of this alpha is
E∝ = 5.69 - 0.241 = 5.449 Mev
Therefore the energies (in MeV) of the two associated alpha particles are; 5.69 MeV and 5.449 Mev
d)
Sketch of the nuclear decay scheme have been uploaded along side this answer.
Answer:
.
Explanation:
<h3>Step One: calculate the coefficients. </h3>Look up the relative atomic mass of these four elements on a modern periodic table:
: approximately
.
: approximately
.
: approximately
.
: approximately
.
The relative atomic mass of an element is numerically equal to the mass (in grams, ,) of one mole of atoms of this element.
For example, the relative atomic mass of is approximately
. Therefore, each mole of
atoms would have a mass of
.
This sample contains of carbon. That would correspond to approximately
of
atoms.
Similarly, for the other three elements:
.
.
.
Hence, the ratio between these elements in this compound would be:
.
In the empirical formula of a compound, the coefficients should represent the smallest possible integer ratio between the number of atoms of these elements.
is indeed the smallest possible integer ratio between the number of atoms of these elements.
Apply the Hill System to arrange these four elements in the empirical formula. In the Hill System:
If carbon, , is present in this compound, then:
- The other elements in this compound will be listed in alphabetical order.
If there is no carbon in this compound, then list all the elements in this compound in alphabetical order.
Both (carbon) and
(hydrogen) are found in this compound. Therefore, the first element in the list would be
. The second would be
, followed by
and then
.
Hence, the empirical formula of this compound would be .