The molarity and normality of 5.7 g of Ca(OH)2 in 450ml 0f solution is calculated as follows
molarity = moles/volume in liters
moles =mass/molar mass
= 5.7g/74g/mol = 0.077moles
molarity = 0.077/450 x1000= 0.17M
Normality = equivalent point x molarity
equivalent point of Ca(OH)2 is 2 since it has two Hydrogen atom
normality is therefore = 0.17 x2 = 0.34 N
First shell from the nucleus can have a maximum of 2 electrons ! hope this helped
Color is the answer Okie :3
The neutralization reaction is:
Ca(OH)2 + 2HBr ---------- CaBr2 + 2H2O
Therefore based on the reaction stoichiometry, the ratio of Ca(OH)2 : HBr = 1:2
Titration data:
Concentration of HBr = 5.00 *10^-2 M
Volume of HBr required = 48.8 ml = 0.0488 L
# moles of HBr = 5.00 *10^-2 moles L-1 * 0.0488 L = 0.00244 moles
Based on the reaction stoichiometry:
Moles of Ca(OH)2 = 1/2 * Moles of HBr = 1/2 * 0.00244 = 0.00122 moles
Volume of Ca(OH)2 taken = 100 ml = 0.1 L
Concentration of Ca(OH)2 = 0.00122/0.1
= 0.0122 M = 1.22 * 10^-2 M
<span>we are given with the equation 3N2H4(l) => 4NH3(g) + N2(g) that shows the decomposition of N2H4 to give ammonia and nitrogen gas. 3 moles of N2H4 forms 4 moles of NH3. When 2.6 moles N2H4 are used, 3.47 moles of ammonia is formed</span>
