Answer:
Cl
Explanation:
The element Cl will have the strongest ionization energy from the given choices. Most non-metals have higher ionization energy compared to metals.
Ionization energy is the energy required to remove the most loosely held electron from the gaseous phase of an atom.
- As you go from left to right on the periodic table, it increases progressive
- From top to bottom, the ionization energy reduces significantly.
- The attractive force between the protons in the nucleus and the electrons plays a very important role.
- In metals, they have very large atomic radius, the attractive force on the outer electrons is very weak.
- This is not the case in non-metals
Explanation:
nuclear fusion yields more energy than nuclear fission and the products of the reaction are not radioactive
37.8 g CH2Br2 X (1 mol CH2Br2 / 173.83 g) = 4.60X10^-3 mol CH2Br2
<span>4.60X10^-3 mol CH2Br2 X (2 mol Br / 1 mol CH2Br2) X 6.02X10^23 atoms/mol = 5.54X10^21 bromine atoms</span>
Answer:
0.29mol/L or 0.29moldm⁻³
Explanation:
Given parameters:
Mass of MgSO₄ = 122g
Volume of solution = 3.5L
Molarity is simply the concentration of substances in a solution.
Molarity = number of moles/ Volume
>>>>To calculate the Molarity of MgSO₄ we find the number of moles using the mass of MgSO₄ given.
Number of moles = mass/ molar mass
Molar mass of MgSO₄:
Atomic masses: Mg = 24g
S = 32g
O = 16g
Molar mass of MgSO₄ = [24 + 32 + (16x4)]g/mol
= (24 + 32 + 64)g/mol
= 120g/mol
Number of moles = 122/120 = 1.02mol
>>>> From the given number of moles we can evaluate the Molarity using this equation:
Molarity = number of moles/ Volume
Molarity of MgSO₄ = 1.02mol/3.5L
= 0.29mol/L
IL = 1dm³
The Molarity of MgSO₄ = 0.29moldm⁻³
<span>Whereas physical properties of a substance tells us about how the substance looks, smells, etc, the chemical properties of a substance basically tells us how the substance will react with other substances. Therefore, knowing the chemical property of a substance tells us how the substance reacts with others.Hope this helps. Let me know if you need additional help!</span><span />
