Answer:
1. 9.57 × 10^-9 moles.
2. 7.38mol
Explanation:
1.) To find the number of moles there are in the number of particles in an atom, we divide the number of particles (nA) by Avagadro's constant (6.02 × 10^23)
Hence, to find the number of moles (n) of Manganese (Mn), we say:
5.76 x 10^15 atoms ÷ 6.02 × 10^23
5.76/6.02 × 10^(15-23)
= 0.957 × 10^-8
= 9.57 × 10^-9 moles.
2.) Mole = mass/molar mass
Molar mass of sodium chloride (NaCl) = 23 + 35.5
= 58.5g/mol
mole = 431.6 g ÷ 58.5g/mol
mole = 7.38mol
The answer that would best complete the given statement above is the term ATOM. <span>The smallest unit which maintains the physical properties of a compound is an ATOM. But for a compound, it would be molecule. Hope this answers your question.</span>
Answer:
235/92U+10n→144/54Xe+90/38Sr+2/10n
Explanation:
- The nuclear reaction for the neutron-induced fission of u−235 to form xe−144 and sr−90 is represented by;
235/92U+10n→144/54Xe+90/38Sr+2/10n
- In nuclear fission reactions a heavy nuclide is split into two light nuclides and is coupled by the release of energy.
Answer:

Explanation:
Hello,
In this case, for the given information, we can compute the rate of disappearance of NO₂ by using the following rate relationship:

Whereas it is multiplied by the the inverse of the stoichiometric coefficient of NO₂ in the reaction that is 2. Moreover, the subscript <em>f</em> is referred to the final condition and the subscript <em>0</em> to the initial condition, thus, we obtain:

Clearly, it turns out negative since the concentration is diminishing due to its consumption.
Regards.
<span>Tf is the freezing point of the solution(the solvent plus solute).
T*f is the freezing point of the pure solvent(without solute)
i is the van't Hoff factor.It is approximately the number of particles in solution that are made for each particle of the solute that is placed into solution.Therefore, for nonelectrolytes, i = 1.
Kf is the freezing point depression constant.For water, Kf = 1.86 Degree C/m, or 1.86 Degree C.kg/mol.
Tf is -1.58 Degree C</span>