Answer: Yes we agree with the student's claim.
Explanation:
When the molecules are present in smaller size, more reactants can react as decreasing the size increases the surface area of the reactants which will enhance the contact of molecules.Hence, more products will form leading to increased rate of reaction.
On increasing the temperature will make more reactant molecules will have sufficient energies to cross the energy barrier and thus the number of effective collisions increases, thus leading to more products and increased rate of reaction.
When the solution is stirred , the molecule's kinetic energy and thus the rate of reaction increases.
Thus smaller size, stirring and increase of temperature will make the solution quickly.
In rubidium oxide - Rb₂O , the ions are Rb⁺ and O²⁻
Rb is a group one element with one valence electron. To become stable it loses its outer electron to gain a complete outer shell.
Electronic configuration of Rb is - 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 5s¹
Once it loses its valence electron the configuration is;
- 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶
The noble gas with this configuration is Krypton - Kr
Oxygen electron configuration is 1s² 2s² 2p⁴
Once it gains 2 electrons the configuration is - 1s² 2s² 3p⁶
The noble gas with this configuration is Neon - Ne
Evaporating.
This is because evaporation occurs for things such as water coming in contact with a heat source that cause molecules to run crazy and begin to go farther and farther apart, becoming steam and evaporation in the air.
Therefore, during the process of evaporation, molecules become less condensed.
Hope this helps!
Answer:
a. 2 HgO(s) ⇒ 2 Hg(l) + O₂(g)
b. 0.957 g
Explanation:
Step 1: Write the balanced equation
2 HgO(s) ⇒ 2 Hg(l) + O₂(g)
Step 2: Convert 130.0 °C to Kelvin
We will use the following expression.
K = °C + 273.15
K = 130.0°C + 273.15
K = 403.2 K
Step 3: Calculate the moles of O₂
We will use the ideal gas equation.
P × V = n × R × T
n = P × V/R × T
n = 1 atm × 0.0730 L/0.0821 atm.L/mol.K × 403.2 K
n = 2.21 × 10⁻³ mol
Step 4: Calculate the moles of HgO that produced 2.21 × 10⁻³ moles of O₂
The molar ratio of HgO to O₂ is 2:1. The moles of HgO required are 2/1 × 2.21 × 10⁻³ mol = 4.42 × 10⁻³ mol.
Step 5: Calculate the mass corresponding to 4.42 × 10⁻³ moles of HgO
The molar mass of HgO is 216.59 g/mol.
4.42 × 10⁻³ mol × 216.59 g/mol = 0.957 g
