Answer:
Explanation:
25.8 ml of .095 N NaOH is needed to neutralise the remaining acid
equivalent of NaOH used = 25.8 x .095 / 1000 = .002451 gm equivalent .
acid remaining = .002451 gm equivalent .
acid initially taken = 100 ml of .1 N / 1000 = . 01 gm equivalent
acid reacted with metal = .01 -.002451 = .007549 gm equivalent
This must have reacted with same gram equivalent of metal oxide
.007549 gm equivalent = .15 gm of metal oxide
1 gm equivalent = 19.87 gm
equivalent weight of metal = 19.87 - equivalent weight of oxygen
= 19.87 - 8 = 11.87 .
1
Answer:
D = 28.2g
Explanation:
Initial temperature of metal (T1) = 155°C
Initial Temperature of calorimeter (T2) = 18.7°C
Final temperature of solution (T3) = 26.4°C
Specific heat capacity of water (C2) = 4.184J/g°C
Specific heat capacity of metal (C1) = 0.444J/g°C
Volume of water = 50.0mL
Assuming no heat loss
Heat energy lost by metal = heat energy gain by water + calorimeter
Heat energy (Q) = MC∇T
M = mass
C = specific heat capacity
∇T = change in temperature
Mass of metal = M1
Mass of water = M2
Density = mass / volume
Mass = density * volume
Density of water = 1g/mL
Mass(M2) = 1 * 50
Mass = 50g
Heat loss by the metal = heat gain by water + calorimeter
M1C1(T1 - T3) = M2C2(T3 - T2)
M1 * 0.444 * (155 - 26.4) = 50 * 4.184 * (26.4 - 18.7)
0.444M1 * 128.6 = 209.2 * 7.7
57.0984M1 = 1610.84
M1 = 1610.84 / 57.0984
M1 = 28.21g
The mass of the metal is 28.21g
Nitrogen lone pair will act as a base,removing H+ from water leaving behind OH- ion.
Why ?
Because N is a better donor than O.
Answer:
1219.5 kj/mol
Explanation:
To reach this result, you must use the formula:
ΔHºrxn = Σn * (BE reactant) - Σn * (BE product)
ΔHºrxn = [1 * (BE C = C) + 2 * (BE C-H) + 5/2 * (BE O = O)] - [4 * (BE C = O) + 2 * (BE O-H).
The BE values are:
BE C = C: 839 kj / mol
BE C-H: 413 Kj / mol
BE O = O: 495 kj / mol
BE C = O = 799 Kj / mol
BE O-H = 463 kj / mol
Now you must replace the values in the above equation, the result of which will be:
ΔHºrxn = [1 * 839 + 2 * (413) + 5/2 * (495)] - [4 * (799) + 2 * (463) = 1219.5 kj/mol
Molality is the number of moles of solute in 1 kg of solvent
number of moles of sucrose - mass of sucrose / molar mass
number of moles of sucrose - 34.2 g / 342.34 g/mol = 0.0999 mol
number of moles in 125 g of water - 0.0999 mol
therefore number of moles in 1000 g - 0.0999 / 125 x 1000 = 0.799 mol/kg
molality of sucrose solution - 0.799 mol/kg
