Compaction and cementation
Positive because it keeps going ok
Answer:
C) the study of the composition of the atom.
Explanation:
A research can be defined as a systematic investigation or careful consideration of study with respect to a particular problem using scientific methods such as collection of data, documenting critical information, analysis of data, and the establishment of facts in order to reach new conclusions.
Similarly, a basic research is an approach to research that's typically theoritical and it's aimed at developing a theory, searching for the truth or gain a better understanding about a phenomenon, subject, or basic laws on nature.
In this context, an example of basic research is the study of the composition of the atom.
An atom can be defined as the smallest unit comprising of matter that forms all chemical elements. Thus, atoms are basically the building blocks of matters and as such determines or defines the structure of a chemical element.
Generally, atoms are typically made up of three distinct particles and these are protons, neutrons and electrons.
Answer:
See explanation and image attached
Explanation:
Yttrium has many isotopes, the lowest mass number of Yttrium is 89Y.
Recall that electron capture converts an electron into a proton and then into a neutron with a consequent emission of a neutrino (v).
In electron capture, the mass number of the daughter nucleus remains the same as that of the parent nucleus while the atomic number of the daughter nucleus is less than that of the parent by one unit.
Answers and Explanation:
a)- The chemical equation for the corresponden equilibrium of Ka1 is:
2. HNO2(aq)⇌H+(aq)+NO−2
Because Ka1 correspond to a dissociation equilibrium. Nitrous acid (HNO₂) losses a proton (H⁺) and gives the monovalent anion NO₂⁻.
b)- The relation between Ka and the free energy change (ΔG) is given by the following equation:
ΔG= ΔGº + RT ln Q
Where T is the temperature (T= 25ºc= 298 K) and R is the gases constant (8.314 J/K.mol)
At the equilibrium: ΔG=0 and Q= Ka. So, we can calculate ΔGº by introducing the value of Ka:
⇒ 0 = ΔGº + RT ln Ka
ΔGº= - RT ln Ka
ΔGº= -8.314 J/K.mol x 298 K x ln (4.5 10⁻⁴)
ΔGº= 19092.8 J/mol
c)- According to the previous demonstation, at equilibrium ΔG= 0.
d)- In a non-equilibrium condition, we have Q which is calculated with the concentrations of products and reactions in a non equilibrium state:
ΔG= ΔGº + RT ln Q
Q= ((H⁺) (NO₂⁻))/(HNO₂)
Q= ( (5.9 10⁻² M) x (6.7 10⁻⁴ M) ) / (0.21 M)
Q= 1.88 10⁻⁴
We know that ΔGº= 19092.8 J/mol, so:
ΔG= ΔGº + RT ln Q
ΔG= 19092.8 J/mol + (8.314 J/K.mol x 298 K x ln (1.88 10⁻⁴)
ΔG= -2162.4 J/mol
Notice that ΔG<0, so the process is spontaneous in that direction.
