Answer:
Freezing point is -2.81°C
Explanation:
34g/342gmol^-1 = 0.0994mol
n = m/mr
Molarity= 0.994/ 0.66 = 1.51M
◇T = -i × m ×Kf
Where ◇T is freezing depression
i= Vant Hoff factor
m = molarity
Kf = freezing content = 1.
860kgmol^-1
◇T =-1 × 1.51 × 1.860 = - 2.81°C
Answer:
True
Explanation:
if there was nothing to avoid of objects pulling towards one another many results may happen
Answer:
the waves have a trough
Explanation:
just took the test on edg.
Answer:
317.6 mL
Explanation:
Step 1: Write the balanced neutralization equation
MgO + 2 HCl ⇒ MgCl₂ + H₂O
Step 2: Calculate the mass corresponding to 640.0 mg of MgO
The molar mass of MgO is 40.30 g/mol. The moles corresponding to 640.0 mg (0.6400 g) of MgO are:
0.6400 g × (1 mol/40.30 g) = 0.01588 mol
Step 3: Calculate the moles of HCl that react with 0.01588 moles of MgO
The molar ratio of MgO to HCl is 1:2. The moles of HCl are 2/1 × 0.01588 mol = 0.03176 mol
Step 4: Calculate the volume of 0.1000 M HCl that contains 0.03176 moles
0.03176 mol × (1 L/0.1000 mol) = 0.3176 L = 317.6 mL
Answer:107.1 g, 124.1 g
Explanation:
The equation of the reaction is;
Al2S3(s) + 6H20(l) ----> 2Al(OH)3(s) + 3H2S(g)
Hence;
For Al2S3
Number of moles= reacting mass/molar mass
Number of moles = 158g/150gmol-1 =1.05 moles
If 1 mole of Al2S3 yields 3 moles of H2S
1.05 moles of Al2S will yield
1.05 × 3/1 = 3.15 moles
Mass of H2S = 3.15moles × 34 gmol-1 = 107.1 g
For water
Number of moles of water = 131g/18gmol-1= 7.3 moles
6 moles of water yields 3 moles of H2S
7.3 moles of water will yield 7.3 × 3/6 = 3.65 moles of H2S
3.65 moles × 34 gmol-1 =124.1 g