Let's divide the three experiments: The experiment with 10.00 mL of water is A), the experiment with 15.00 mL is B), and the experiment with 25.00 mL is C).
- (1) Now let's calculate the experimental density of each experiment. Density (ρ) is equal to the mass divided by the volume, thus:
- (2)To calculate the average density, we add each density and divide the result by the number of experiments (in this case 3):
- (3) The percent error is calculated by dividing the absolute value of the substraction of the theorethical and experimental values, by the theoretical value, times 100:
%error=
%error=
%error=2.44 %
Answer
pH=8.5414
Procedure
The Henderson–Hasselbalch equation relates the pH of a chemical solution of a weak acid to the numerical value of the acid dissociation constant, Kₐ. In this equation, [HA] and [A⁻] refer to the equilibrium concentrations of the conjugate acid-base pair used to create the buffer solution.
pH = pKa + log₁₀ ([A⁻] / [HA])
Where
pH = acidity of a buffer solution
pKa = negative logarithm of Ka
Ka =acid disassociation constant
[HA]= concentration of an acid
[A⁻]= concentration of conjugate base
First, calculate the pKa
pKa=-log₁₀(Ka)= 8.6383
Then use the equation to get the pH (in this case the acid is HBrO)
Answer:
4190.22 L = 4.19 m³.
Explanation:
- For the balanced reaction:
<em>2P₂ + 5O₂ ⇄ 2P₂O₅. </em>
It is clear that 2 mol of P₂ react with <em>5 mol of O₂ </em>to produce <em>2 mol of P₂O₅.</em>
- Firstly, we need to calculate the no. of moles of 6.92 kilograms of P₂O₅ produced through the reaction:
no. of moles of P₂O₅ = mass/molar mass = (6920 g)/(283.88 g/mol) = 24.38 mol.
- Now, we can find the no. of moles of O₂ is needed to produce the proposed amount of P₂O₅:
<u><em>Using cross multiplication:</em></u>
5 mol of O₂ is needed to produce → 2 mol of P₂O₅, from stichiometry.
??? mol of O₂ is needed to produce → 24.38 mol of P₂O₅.
∴ The no. of moles of O₂ needed = (5 mol)(24.38 mol)/(2 mol) = 60.95 mol.
- Finally, we can get the volume of oxygen using the general law of ideal gas:<em> PV = nRT.</em>
where, P is the pressure of the gas in atm (P = 606.1 mm Hg/760 = 0.8 atm).
V is the volume of the gas in L (V = ??? L).
n is the no. of moles of the gas in mol (n = 60.95 mol).
R is the general gas constant (R = 0.0821 L.atm/mol.K),
T is the temperature of the gas in K (396.90°C + 273 = 669.9 K).
∴ V of oxygen needed = nRT/P = (60.95 mol)(0.0821 L.atm/mol.K)(669.9 K)/(0.8 atm) = 4190.22 L/1000 = 4.19 m³.
Mass 1 + %abundance of first isotope + Mass 2 + %abundance of second isotope
/ 100
This is RAM.
