Movement of molecules from an area of higher concentration to one of lower concentration is called Diffusion
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
Answer:
A molecular compound is usually composed of two or more nonmetal elements. Molecular compounds are named with the first element first and then the second element by using the stem of the element name plus the suffix -ide.
Molecules are cooler so they bocme more dense.. move very slow, have less kinetic energy and are not in random motion.
5H2O2 + 2KMnO4<span>+ 3H2SO4 = 5O2 + 2MnSO4 + 8H2O + K2SO4
0,145 moles of KMnO4----------in--------1000ml
x moles of KMnO4---------------in------------46ml
x = 0,00667 moles of KMnO4
according to the reaction:
2 moles of KMnO4------------------5 moles of H2O2
0,00667 moles of KMnO4----------------x
x = 0,01668 moles of H2O2
0,01668 moles of H2O2---------in-----------50ml
x moles of H2O2--------------------in----------1000ml
<u>x = 0,334 mol/L H2O2</u></span>
