Answer:
Alkali metals
Explanation:
All alkali metals have 1 valence electron, and they want to lose it, because if they do, they will have 8 valence electrons, which will make them inert.
When an atom lose an electron, the atom becomes a cation, because the atom will have one less electrons than protons.
I hope this helps
Answer:
a) <u>Balanced Equation</u>
2NaCl (aq) + (NH₄)₂CO₃ (aq) → Na₂CO₃ (aq) + 2NH₄Cl (aq)
b) <u>Total Ionic Equation</u>
2Na⁺ (aq) + 2Cl⁻ (aq) + 2NH₄⁺ (aq) + CO₃ ²⁻ (aq) → 2Na⁺ (aq) + CO₃²⁻ (aq) + 2NH₄⁺ (aq) + 2Cl⁻ (aq)
c) <u>Net Ionic Equation</u>
All are spectator ions, so no net ionic equation.
<u>Examples of energy</u>: Heat, Electrical, Mechanical, Nuclear, Chemical, Kinetic and Potential.
<u>Non-Examples of energy</u>: Ideas, matter, water, paper, people and coffee.
Answer:
0.7561 g.
Explanation:
- The hydrogen than can be prepared from Al according to the balanced equation:
<em>2Al + 6HCl → 2AlCl₃ + 3H₂,</em>
It is clear that 2.0 moles of Al react with 6.0 mole of HCl to produce 2.0 moles of AlCl₃ and 3.0 mole of H₂.
- Firstly, we need to calculate the no. of moles of (6.8 g) of Al:
no. of moles of Al = mass/atomic mass = (6.8 g)/(26.98 g/mol) = 0.252 mol.
<em>Using cross multiplication:</em>
2.0 mol of Al produce → 3.0 mol of H₂, from stichiometry.
0.252 mol of Al need to react → ??? mol of H₂.
∴ the no. of moles of H₂ that can be prepared from 6.80 g of aluminum = (3.0 mol)(0.252 mol)/(2.0 mol) = 0.3781 mol.
- Now, we can get the mass of H₂ that can be prepared from 6.80 g of aluminum:
mass of H₂ = (no. of moles)(molar mass) = (0.3781 mol)(2.0 g/mol) = 0.7561 g.
Hey :)
molarity<span> is affected by </span>temperature<span> because morality is simply a measure on the volume of a particular </span>solution<span>, and the volume of a substance will be negatively or positively affected by changes in </span>temperature. In theory, volume will increase<span> when </span>temperature increases<span>, a reciprocal effect of </span><span>molarity</span>
