Answer:
81.26% is the percent yield
Explanation:
Based on the reaction:
CaCl₂ + Na₂CO₃ → 2NaCl + CaCO₃
<em>Where 1 mole of CaCl₂ in excess of sodium carbonate produces 1 mole of calcium carbonate.</em>
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To solve this question we must find the moles of CaCl2 added = Moles CaCO₃ produced (Theoretical yield). The percent yield is:
Actual yield (0.366g) / Theoretical yield * 100
<em>Moles CaCl₂ = Moles CaCO₃:</em>
0.0500L * (0.0900moles / L) = 0.00450 moles of CaCO₃
<em>Theoretical mass -Molar mass CaCO₃ = 100.09g/mol-:</em>
0.00450 moles of CaCO₃ * (100.09g / mol) = 0.450g of CaCO₃
Percent yield = 0.366g / 0.450g * 100
81.26% is the percent yield
Answer is: 230 g.
Chemical reaction: P₄ + 5O₂ → 2P₂O₅.
m(P₄) = 100 g.
M(P₄) = 4 · 31 g/mol = 124 g/mol.
n(P₄) = m(P₄) ÷ M(P₄) = 100g ÷ 124g/mol = 0,806 mol.
From reaction: n(P₄) : n(P₂O5) = 1 : 2.
n(P₂O₅) = 1,612 mol.
m(P₂O₅) = 1,612 mol · 142g/mol = 230g.
M - molar mass.
n - amount of substance.
To determine the molar mass, you need to get the atomic mass of the molecule. To do this, check the periodic table for the atomic mass or average atomic weight of each element.
Mg = 24.305 x 1 = 24.305 amu
O = 15.9994 x 2 =31.9988 amu
H = 1.0079 x 2 = 2.0158 amu
Then, add all the components to get the atomic mass of the molecule.
24.305 amu + 31.9988 amu + 2.0158 amu = 58.3196 amu
The atomic mass is just equivalent to its molar mass.
So, the molar mass of Magnesium hydroxide (Mg(OH)2) is 58.3196 g/mol.
Answer:
b) It produces electrical current spontaneously.
Explanation:
Cells capable of converting chemical energy to electrical energy and vice versa are termed Electrochemical cells. There are two types of electrochemical cells viz: Galvanic or Voltaic cells and Electrolytic cells. Voltaic cell is an elctrochemical cell capable of generating electrical energy from the chemical reaction occuring in it.
The voltaic cell uses spontaneous reduction-oxidation (redox) reactions to generate ions in a half cell that causes electric currents to flow. An half cell is a part of the galvanic cell where either oxidation or reduction reaction is taking place. Hence, the spontaneous production of electric currents is true about Voltaic/Galvanic cells.
This is what i have found i hope this helps
