B. solvent
solute is the substance being dissolved
Answer:
a) ammonium ion
b) amide ion
Explanation:
The order of decreasing bond angles of the three nitrogen species; ammonium ion, ammonia and amide ion is NH4+ >NH3> NH2-. Next we need to rationalize this order of decreasing bond angles from the valence shell electron pair repulsion (VSEPR) theory perspective.
First we must realize that all three nitrogen species contain a central sp3 hybridized carbon atom. This means that a tetrahedral geometry is ideally expected. Recall that the presence of lone pairs distorts molecular structures from the expected geometry based on VSEPR theory.
The amide ion contains two lone pairs of electrons. Remember that the presence of lone pairs causes greater repulsion than bond pairs on the outermost shell of the central atom. Hence, the amide ion has the least H-N-H bond angle of about 105°.
The ammonia molecule contains one lone pair, the repulsion caused by one lone pair is definitely bless than that caused by two lone pairs of electrons hence the bond angle of the H-N-H bond in ammonia is 107°.
The ammonium ion contains four bond pairs and no lone pair of electrons on the outermost nitrogen atom. Hence we expect a perfect tetrahedron with bond angle of 109°.
Explanation:
The reaction given is;
TiCl4 + H2O --> TiO2 + HCl
The reaction is not balanced, upon balancing it is given as;
TiCl4 + 2H2O → TiO2 + 4HCl
a. How many moles of H2O are needed to react with 6.50 moles of TiCl4?
From the reaction;
1 mol of TiCl4 requires 2 mol of H2O
6.50 mol of TiCl4 would require x mol of H2O
1 = 2
6.5 = x
x = 6.5 * 2 / 1 = 13.0 mol
b. How many moles of HCl are formed when 8.44 moles of TiCl4 react?
From the equation of the reaction;
1 mol of TiCl4 reacts to form 4 mol of HCl
8.44 mol of TiCl4 reacts to form x mol of HCl
1 = 4
8.44 = x
x = 8.44 * 4 / 1 = 33.76 mol
Answer:
The temperature at which the liquid vapor pressure will be 0.2 atm = 167.22 °C
Explanation:
Here we make use of the Clausius-Clapeyron equation;
Where:
P₁ = 1 atm =The substance vapor pressure at temperature T₁ = 282°C = 555.15 K
P₂ = 0.2 atm = The substance vapor pressure at temperature T₂
= The heat of vaporization = 28.5 kJ/mol
R = The universal gas constant = 8.314 J/K·mol
Plugging in the above values in the Clausius-Clapeyron equation, we have;
T₂ = 440.37 K
To convert to Celsius degree temperature, we subtract 273.15 as follows
T₂ in °C = 440.37 - 273.15 = 167.22 °C
Therefore, the temperature at which the liquid vapor pressure will be 0.2 atm = 167.22 °C.
