Answer:
different isotopes of na element
The reason that some of the elements of period three and beyond are steady in spite of not sticking to the octet rule is due to the fact of possessing the tendency of forming large size, and a tendency of making more than four bonds. For example, sulfur, it belongs to period 3 and is big enough to hold six fluorine atoms as can be seen in the molecule SF₆, while the second period of an element like nitrogen may not be big to comprise 6 fluorine atoms.
The existence of unoccupied d orbitals are accessible for bonding for period 3 elements and beyond, the size plays a prime function than the tendency to produce more bonds. Hence, the suggestion of the second friend is correct.
Answer:
The balanced equation tells us that 1 mole of Zn will produce 1 mole of H2.
1.566 g Zn x (1 mole Zn / 65.38 g Zn) = 0.02395 moles Zn
0.02395 moles Zn x (1 mole H2 / 1 mole Zn) = 0.02395 moles H2 produced
Now use the ideal gas law to find the volume V.
P = 733 mmHg x (1 atm / 760 atm) = 0.964 atm
T = 21 C + 273 = 294 K
PV = nRT
V = nRT/ P = (0.02395 moles H2)(0.0821 L atm / K mole)(294 K) / (0.964 atm) = 0.600 L
B. carbon-13 is not an allotrope of Carbon.
Allotropes<span> are elements on the periodic table that have more than one crystalline form. </span>Isotopes<span> are atoms of the same element with the same atomic number but have a different mass number.
C-13 is an isotope of carbon, not an allotrope.</span>
Answer:
25.7 kJ/mol
Explanation:
There are two heats involved.
heat of solution of NH₄NO₃ + heat from water = 0
q₁ + q₂ = 0
n = moles of NH₄NO₃ = 8.00 g NH₄NO₃ × 1 mol NH₄NO₃/80.0 g NH₄NO₃
∴ n = 0.100 mol NH₄NO₃
q₁ = n * ΔHsoln = 0.100 mol * ΔHsoln
m = mass of solution = 1000.0 g + 8.00 g = 1008.0 g
q₂ = mcΔT = 58.0 g × 4.184 J°C⁻¹ g⁻¹ × ((20.39-21)°C) = -2570.19 J
q₁ + q₂ = 0.100 mol ×ΔHsoln – 2570.19 J = 0
ΔHsoln = +2570.19 J /0.100 mol = +25702 J/mol = +25.7 kJ/mol
