Answer:
18.0 g of mercury (11) oxide decomposes to produce 9.0 grams of mercury
Explanation:
Mercury oxide has molar mass of 216.6 g/ mol. It gas a molecular formula of HgO.
The decomposition of mercury oxide is given by the chemical equation below:
2HgO ----> 2Hg + O₂
2 moles of HgO decomposes to produce 1 mole of Hg
2 moles of HgO has a mass of 433.2 g
433.2 g of HgO produces 216.6 g of Hg
18.0 of HgO will produce 18 × 216.6/433.2 g of Hg = 9.0 g of Hg
Therefore, 18.0 g of mercury (11) oxide decomposes to produce 9.0 grams of mercury
D. The number increases and then decreases for noble gases
175.8 g NaCl to moles:
(175.8 g)/(58.44 g/mol) = 3.008 mol NaCl
Molarity = moles of solute/volume of solution in liters
(3.008 mol NaCl)/(1.5 L) = 2.0 M.
The molarity of this solution would be 2.0 M.
Answer:
14.3mL you require to reach the half-equivalence point
Explanation:
A strong acid as HClO₄ reacts with a weak base as CH₃CH₂NH₂, thus:
CH₃CH₂NH₂ + HClO₄ → CH₃CH₂NH₃⁺ + ClO₄⁻
As the reaction is 1:1, to reach the equivalence point you require to add the moles of HClO₄ equal to moles CH₃CH₂NH₂ you add originally. Also, half-equivalence point requires to add half-moles of CH₃CH₂NH₂ you add originally.
Initial moles of CH₃CH₂NH₂ are:
20.8mL = 0.0208L × (0.51mol CH₃CH₂NH₂ / 1L) =
0.0106moles CH₃CH₂NH₂
To reach the half-equivalence point you require:
0.0106moles ÷ 2 = 0.005304 moles HClO₄
As concentration of HClO₄ is 0.37M, volume you require to add 0.005304moles is:
0.005304 moles HClO₄ ₓ (1L / 0.37mol) = 0.0143L =
<h3> 14.3mL you require to reach the half-equivalence point</h3>
Because size of an atom INCREASES from top to bottom and DECREASES from left to right on the periodic table. If to elements are in the same period they can't increase in size based on the up or down trend. Instead, notice that Alkali Metals are LEFT of Alkaline Earth Metals. Therefore, the Alkali Metal will be larger than the Alkaline Earth Metals.
