<span>6CO</span>₂ +<span> 6H</span>₂O +<span> energy → C</span>₆H₁₂O₆<span> 6O</span>₂
Density is defined as mass per unit volume so even when you cut an object in half unit volume does not change so each part would have a different density even if it’s cut into the same pieces
Answer:
Explanation:
mole of NaOH present = molarity x volume
= 1.0 X 0.05 = 0.05 mole
<em>Recommended mole of HCl </em>= 1.1 x 0.05 = 0.055
<em>Mole of HCl carelessly added by Jacob </em>= 1.1 x 0.04 = 0.044
From the equation of reaction:
HCl + NaOH ----> NaCl + H2O
The ratio of mole of HCl to that of NaOH for a complete neutralization reaction is 1:1. However, the recommended mole of HCl (0.055 mole) is more than the mole of NaOH (0.05 mole). <u>Hence, the recommended endpoint of the reaction is supposed to be acidic.</u>
The mole of HCl added by Jacob (0.044) is short of the recommended amount (0.055) and also short of the amount required for a neutral endpoint (0.05). <u>This means that the endpoint will have an excess amount of NaOH and as such, basic instead of the desired acidic endpoint.</u>
Answer:
1) It is the number of particles in one mole of substance. It is an arbitrary definition to denote an amount of substance, not by its weight but by its number of molecules. Why? because that is convenient when doing certain calculations involving molecule formulas.
2) 1 mole of substance contains exactly as many particles as there are atoms in 12 grams of carbon-12.
3) The unified atomic mass unit is 1/12 of the mass of 1 carbon-12 atom.
So you see all three concepts are linked together.
Answer:
Percent yield = 90.9%
Explanation:
Given data:
Mass of CaCO₃ = 50.0 g
Mass of CO₂ produced = 20.0 g
Percent yield = ?
Solution:
Chemical equation:
CaCO₃ → CaO + CO₂
Number of moles of CaCO₃:
Number of moles = mass/molar mass
Number of moles = 50.0 g/ 100.1 g/mol
Number of moles = 0.5 mol
Now we will compare the moles of CO₂ with CaCO₃.
CaCO₃ : CO₂
1 : 1
0.5 : 0.5
Mass of CO₂: Theoretical yield
Mass = number of moles × molar mass
Mass = 0.5 mol × 44 g/mol
Mass = 22 g
Percent yield:
Percent yield = ( actual yield / theoretical yield ) × 100
Percent yield = (20.0 g/ 22.0 g) × 100
Percent yield = 0.909 × 100
Percent yield = 90.9%
