The arrangement of valence electrons in metals can be best described by the
A sea of electrons.
Answer: 4.05 mol O2, 15.36 mol H2O
Explanation:
I can answer each question individually if you post them individually.
2a. 3.24 mol NH3 * (5 mol O2 / 4 mol NH3) = 4.05 mol O2
2b. 12.8 mol O2 ( 6mol H2O/ 5 mol O2) = 15.36 mol H2O
Essentially what I did was dimensional analysis. Multiplying in a way that the units cancel out so the only thing remains is what each question asks for.
Answer is: the name of the unknown oxyanion is hypobromite.
<span>
M(NaBrO</span>ₓ) = 118.9 g/mol; <span>molecular weight of an ionic compound.
</span>M(Na) = 23 g/mol; atomic weight of sodium.
M(Br) = 79.9 g/mol; atomic weight of bromine.
M(O) = 16 g/mol; atomic weight of oxygen.
M(Na) + M(Br) + M(O) = 23 g/mol + 79.9 g/mol + 16 g/mol.
M(Na) + M(Br) + M(O) = 118.9 g/mol, so compound is NaBrO and oxyanion is BrO⁻.
Answer: A minor but very important component of the atmosphere, carbon dioxide is released through natural processes such as respiration and volcano eruptions and through human activities such as deforestation, land use changes, and burning fossil fuels.
Explanation:
The balanced equation for the above neutralisation reaction is as follows;
2KOH + H₂SO₄ --> K₂SO₄ + 2H₂O
stoichiometry of KOH to H₂SO₄ is 2:1
neutralisation is the reaction between H⁺ ions and OH⁻ ions to form water which is neutral
number of KOH moles - 1.56 mol
2 mol of KOH require 1 mol of H₂SO₄ for neutralisation
therefore 1.56 mol of KOH require - 1/2 x 1.56 mol = 0.78 mol
0.78 mol of H₂SO₄ are required for neutralisation
