Answer:
Lithium
Explanation:
Alkali metals are group of metals which are present in first group of periodic table. As we know atomic number is equal to number of protons contained by a particular element. Therefore, the alkali metals along with there number of protons are listed below;
Alkali Metal Number of Protons
Lithium 3
Sodium 11
Potassium 19
Rubidium 37
Cesium 55
Francium 87
Hence, it is cleared from above table that Lithium is having fewer protons than 10.
Answer:
Potassium
General Formulas and Concepts:
<u>Chem</u>
- Reading a Periodic Table
- Periodic Trends
- Ionization Energy - energy required to remove an electron from a given element
- Coulomb's Law
- Shielding Effect
- Z-effective and Forces of Attraction
Explanation:
The Periodic Trend for 1st Ionization Energy is increasing up and to the right. That means He would have the highest I.E and therefore take the most amount of energy to remove an electron.
Potassium and Gallium are both in Period 4. Potassium is element 19 and Gallium is element 31.
Potassium's electron configuration is [Ne] 4s¹ and Gallium's electron configurations is [Ne] 4s²3d¹⁰4p¹. Since both are in Period 4, they have the same number of core e⁻. Therefore, the shielding effect is the same.
However, since Gallium is element 31, it has 31 protons compared to Potassium, which is element 19 and has 19 protons. Gallium would have a greater Zeff than Potassium as it has more protons. Therefore, the FOA between the electrons and nucleus of Ga is much stronger than that of K. Thus, Ga requires <em>more</em> energy to overcome those FOA to remove the 4p¹ e⁻. Since K has less protons, it will have a smaller Zeff and thus less FOA between the e⁻ and nucleus, requiring <em>less</em> energy to remove the 4s¹ e⁻.
Responda:
+ 0,9kJ / mol
Explicação:
Dados os calores de combustão do enxofre monoclínico e enxofre rômbico como - 297,2 kJ / mol e - 296,8 kJ / mol, respectivamente para a variação na transformação de 1 mol de enxofre rômbico em enxofre monoclínico conforme mostrado pela equação;
S (mon.) + O2 (g) -> SO2 (g)
Uma vez que são todos 1 mol cada, a mudança na entalpia será expressa como ∆H = ∆H2-∆H1
Dado ∆H2 = -296,8kJ / mol
∆H1 = -297,2kJ / mol
∆H = -296,8 - (- 297,2)
∆H = -296,8 + 297,2
∆H = 297,2-296,8
∆H = + 0,9kJ / mol
Portanto, a mudança na entalpia da equação é + 0,9kJ / mol
Answer:
Synthesis reaction
Explanation:
This is a synthesis reaction. When you combine sodium metal (Na) with chlorine gas (Cl), you form sodium chloride or kosher salt (NaCl).
<em>Hope this explanation helps!</em>
Answer:
ΔT = 20.06 °C
Explanation:
The equation used for this problem is as follow,
Q = m Cp ΔT ----- (1)
Where;
Q = Heat = 1.17 kJ = 1170 J
m = mass = 24.1 g
Cp = Specific Heat Capacity = 2.42 J.g⁻¹.°C⁻¹
ΔT = Change in Temperature = <u>??</u>
Solving eq. 1 for ΔT,
ΔT = Q / m Cp
Putting values,
ΔT = 1170 J / 24.1 g × 2.42 J.g⁻¹.°C⁻¹
ΔT = 20.06 °C
