Answer:
691.7 mmHg is the resulting pressure
Explanation:
Tha Gay-Lussac's law states that the pressure of a gas is directly proportional to its absolute temperature under constant volume. The equation is:
P1T2 = P2T1
<em>Where P is pressure and T asbolute temperature of 1, initial state and 2, final state of the gas.</em>
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Computing the values of the problem:
T1 = 273 + 25 = 298K
P1 = 755.0mmHg
T2 = 273 + 0 = 273K
P2 = ?
755.0mmHg*273K = P2*298K
<h3>691.7 mmHg is the resulting pressure</h3>
Answer:
T°fussion of solution is -18°C
Explanation:
We have to involve two colligative properties to solve this. Let's imagine that the solute is non electrolytic, so i = 1
First of all, we apply boiling point elevation
ΔT = Kb . m . i
ΔT = Boiling T° of solution - Boiling T° of pure solvent
Kb = ebuliloscopic constant
105°C - 100° = 0.512 °C kg/mol . m . 1
5°C / 0.512 °C mol/kg = m
9.7 mol/kg = m
Now that we have the molality we can apply, the Freezing point depression.
ΔT = Kf . m . i
Kf = cryoscopic constant
0° - (T°fussion of solution) = 1.86 °C/m . 9.76 m . 1
- (1.86°C /m . 9.7 m) = T°fussion of solution
- 18°C = T°fussion of solution
Answer:
a group of unicellular microorganisms
Answer:
4.00 is the pH of the mixture
Explanation:
The ethyl amine reacts with HNO3 as follows:
C2H5NH2 + HNO3 → C2H5NH3⁺ + NO3⁻
To solve this question we need to find the moles of ethyl amine and the moles of HNO3:
<em>Moles C2H5NH2:</em>
0.0500L * (0.100mol/L) = 0.00500 moles ethyl amine
<em>Moles HNO3:</em>
0.201L * (0.025mol/L) = 0.005025 moles HNO3
That means HNO3 is in excess. The moles in excess are:
0.005025 moles HNO3 - 0.00500 moles ethyl amine =
2.5x10⁻⁵ moles HNO₃
In 50 + 201mL = 251mL = 0.251L:
2.5x10⁻⁵ moles HNO₃ / 0.251L = 9.96x10⁻⁵M = [H+]
As pH = -log [H+]
pH = -log 9.96x10⁻⁵M
pH = 4.00 is the pH of the mixture
