1_ Al in 1.200 g impure [NH4Al (SO4) 2] (237 g / mol) is precipitated with NH3 as Al2O3 × H2O. The precipitate is filtered Al2O3 (102 g / mol) 0.178 g% What is the amount of Al (27 g / mol)?
2_ 100 ml 0.25 M NH3, 50ml 0.2 M HCl
(Ka=1.75×10^-5) pH= ?
3_ 0.250 M CH3COOH , 0.125 M NaOH 50 ml (CH3COOH Ka= 1.75×10^-5) pH=?
4_ Of CO2 (44 g/mol) (wanted) in CaCO (100 g/mol) what is the gravimetric factor?
sorry dude i'm not good at math
4_ Of CO2 (44 g/mol) (wanted) in CaCO (100 g/mol) what is the gravimetric factor?
Interesting. I thought they stopped teaching gravimetric factors years ago. Also I thought that was a mistake. I assume you meant CaCO3 and not CaCO.
The factor is (molar mass CO2/molar mass CaCO3) = 44/100 = ?
3_ 0.250 M CH3COOH , 0.125 M NaOH 50 ml (CH3COOH Ka= 1.75×10^-5) pH=?
mols CH3COOH = 0.250
CH3COOH + NaOH --> CH3COONa + H2O
How much CH3COOH? Gotta know that.
#2. You have the Ka for NH3 wrong.
#1. I can't make sense of the problem the way you've presented it.
1. To determine the amount of Al present in the precipitate, we need to calculate the moles of Al2O3 in the precipitate and then convert it to moles of Al.
First, calculate the moles of Al2O3 in the precipitate:
Mass of Al2O3 = 0.178 g
Molar mass of Al2O3 = 102 g/mol
Moles of Al2O3 = mass / molar mass = 0.178 g / 102 g/mol
Next, convert moles of Al2O3 to moles of Al:
In the balanced equation of the reaction, we have 2 moles of Al2O3 for every 2 moles of Al.
So, the moles of Al = Moles of Al2O3 = moles of Al2O3
Finally, calculate the amount of Al:
Molar mass of Al = 27 g/mol
Amount of Al = moles of Al x molar mass of Al = moles of Al2O3 x molar mass of Al
2. To determine the pH of the resulting solution, we need to calculate the final concentrations of NH4+ and NH3 after the reaction between NH3 and HCl.
Given:
Volume of NH3 = 100 ml
Concentration of NH3 = 0.25 M
Volume of HCl = 50 ml
Concentration of HCl = 0.2 M
Ka for NH4+ = 1.75 × 10^-5
First, calculate the moles of NH3 and HCl:
Moles of NH3 = concentration x volume = 0.25 M x 0.1 L
Moles of HCl = concentration x volume = 0.2 M x 0.05 L
The reaction between NH3 and HCl is:
NH3 + HCl → NH4+ + Cl-
Since the reactants have a 1:1 stoichiometric ratio, the amount of NH4+ formed will be equal to the amount of NH3 reacted.
Next, calculate the concentration of NH4+:
Concentration of NH4+ = Moles of NH4+ / Total volume of solution
Assuming the volumes of NH3 and HCl are additive:
Total volume of solution = Volume of NH3 + Volume of HCl = 100 ml + 50 ml
Finally, calculate the pH using the Henderson-Hasselbalch equation:
pH = pKa + log10([A-] / [HA])
Since NH3 is a weak base and NH4+ is its conjugate acid:
[HA] = Concentration of NH4+
[A-] = Concentration of NH3 (which is equal to the concentration of NH4+)
3. To determine the pH of the resulting solution, we need to calculate the final concentrations of CH3COOH (acetic acid) and CH3COO- (acetate) after the reaction between CH3COOH and NaOH.
Given:
Volume of CH3COOH = 50 ml
Concentration of CH3COOH = 0.250 M
Volume of NaOH = 50 ml
Concentration of NaOH = 0.125 M
Ka for CH3COOH = 1.75 × 10^-5
First, calculate the moles of CH3COOH and NaOH:
Moles of CH3COOH = concentration x volume = 0.250 M x 0.05 L
Moles of NaOH = concentration x volume = 0.125 M x 0.05 L
The reaction between CH3COOH and NaOH is:
CH3COOH + NaOH → CH3COO- + H2O
Since the reactants have a 1:1 stoichiometric ratio, the amount of CH3COO- formed will be equal to the amount of CH3COOH reacted.
Next, calculate the concentration of CH3COO-:
Concentration of CH3COO- = Moles of CH3COO- / Total volume of solution
Assuming the volumes of CH3COOH and NaOH are additive:
Total volume of solution = Volume of CH3COOH + Volume of NaOH = 50 ml + 50 ml
Finally, calculate the pH using the Henderson-Hasselbalch equation:
pH = pKa + log10([A-] / [HA])
Since CH3COOH is a weak acid and CH3COO- is its conjugate base:
[HA] = Concentration of CH3COOH
[A-] = Concentration of CH3COO- (which is equal to the concentration of CH3COOH)
4. To calculate the gravimetric factor, we need to determine the molar ratio between CaCO3 and CO2.
Given:
Molar mass of CO2 = 44 g/mol
Molar mass of CaCO3 = 100 g/mol
The balanced equation for the reaction is:
CaCO3 → CaO + CO2
From the balanced equation, we can see that 1 mole of CaCO3 produces 1 mole of CO2. Therefore, the gravimetric factor is 1.
Note: The gravimetric factor represents the molar ratio between the substance you are interested in (in this case, CO2) and the substance you have given information about (in this case, CaCO3).
1) To determine the amount of Al (27 g/mol) in the precipitate, we need to first calculate the moles of Al2O3.
The mass of Al2O3 is given as 0.178 g%. To convert this to grams, we need to assume a total weight for the precipitate. Let's assume the total weight of the precipitate is 100 g.
0.178 g% of 100 g = 0.178 g
Next, we calculate the moles of Al2O3:
Molar mass of Al2O3 = 102 g/mol
Moles of Al2O3 = mass of Al2O3 / molar mass of Al2O3
Moles of Al2O3 = 0.178 g / 102 g/mol
Since the balanced equation for the precipitation reaction states that 1 mole of Al2O3 corresponds to 2 moles of Al, we can use the stoichiometry to calculate the moles of Al:
Moles of Al = (moles of Al2O3) * 2
Finally, we calculate the mass of Al:
Mass of Al = Moles of Al * molar mass of Al
2) To find the pH of the solution, we need to consider the reaction between NH3 (a weak base) and HCl (a strong acid). The reaction between the two involves the formation of NH4+ (ammonium) and Cl- (chloride) ions.
First, we need to determine which of the two reactants will be completely consumed. In this case, HCl is present in limited quantity (50 ml * 0.2 M = 10 mmol), while NH3 is present in excess (100 ml * 0.25 M = 25 mmol). So, HCl will be completely consumed in the reaction.
Next, we calculate the number of moles of NH3 remaining after the reaction with HCl:
Moles of NH3 = initial moles of NH3 - moles of HCl reacted
Moles of NH3 = 25 mmol - 10 mmol
Now, since NH3 is a base, it will react with water to form NH4+ and OH- ions. This OH- concentration can be used to calculate pOH and then convert it to pH using the equation pH = 14 - pOH.
Lastly, we need to account for the dissociation of water. The concentration of OH- ions is related to the concentration of H+ ions through the equilibrium constant for water, Kw = [H+][OH-] = 1.0 x 10^-14.
We can use the equation Ka * Kb = Kw to determine the Kb value for NH3 using the given Ka value for HCl. Then, we can use this value to calculate the OH- concentration and pH of the solution.
3) The pH of an acid-base solution can be determined by comparing the concentration of the acid (CH3COOH) with the concentration of the conjugate base (CH3COO-) after reaction with NaOH.
First, we need to determine which of the two reactants (CH3COOH or NaOH) will be completely consumed. In this case, NaOH is present in limited quantity (0.125 M * 50 ml = 6.25 mmol), while CH3COOH is present in excess (0.250 M * 50 ml = 12.5 mmol). So, NaOH will be completely consumed in the reaction.
Next, we calculate the number of moles of CH3COOH remaining after the reaction with NaOH:
Moles of CH3COOH = initial moles of CH3COOH - moles of NaOH reacted
Moles of CH3COOH = 12.5 mmol - 6.25 mmol
Now, since CH3COOH is a weak acid, it will partially dissociate in water, resulting in the formation of H+ and CH3COO- ions. The concentration of H+ ions can be used to calculate pH.
To determine the pH, we first calculate the concentration of H+ ions using the dissociation constant Ka for CH3COOH. Then, we can use the equation pH = -log[H+] to find the pH of the solution.
4) To calculate the gravimetric factor for the conversion between CO2 (44 g/mol) and CaCO (100 g/mol), we need to determine the ratio of their molar masses.
Gravimetric factor = molar mass of CO2 / molar mass of CaCO
By dividing the molar mass of CO2 (44 g/mol) by the molar mass of CaCO (100 g/mol), we get the gravimetric factor.