Answer : The work done by the system is, 2.2722 J
Explanation :
The expression used for work done in reversible isothermal expansion will be,
where,
w = work done = ?
n = number of moles of gas = 0.00100 mole
R = gas constant = 8.314 J/mole K
T = temperature of gas = 
= initial volume of gas = 25 mL
= final volume of gas = 75 mL
Now put all the given values in the above formula, we get:
Therefore, the work done by the system is, 2.2722 J
Answer:
Explanation:
a )
m = m₀ 
m is mass after time t . original mass is m₀ , λ is disintegration constant
λ = .693 / half life
= .693 / 1590
= .0004358
m = m₀ 
b )
m = 50 x 
= 40.21 mg .
c )
40 = 50 
.8 =
= 1.25
.0004358 t = .22314
t = 512 years .
Answer:
34.9 g/mol is the molar mass for this solute
Explanation:
Formula for boiling point elevation: ΔT = Kb . m . i
ΔT = Temperatures 's difference between pure solvent and solution → 0.899°C
Kb = Ebullioscopic constant → 0.511°C/m
m = molality (moles of solute/1kg of solvent)
i = 2 → The solute is a strong electrolyte that ionizes into 2 ions
For example: AB ⇒ A⁺ + B⁻
Let's replace → 0.899°C = 0.511 °C/m . m . 2
0.899°C / 0.511 m/°C . 2 = m → 0.879 molal
This moles corresponds to 1 kg of solvent. Let's determine the molar mass
Molar mass (g/mol) → 30.76 g / 0.879 mol = 34.9 g/mol
The symbol for the hydroxide ion is OH-
Answer:
0.185M sulfuric acid
Explanation:
Based on the reaction:
H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O
<em>1 mole of sulfuric acid reacts with 2 moles of KOH</em>
Initial moles of H₂SO₄ and KOH are:
H₂SO₄: 0.750L ₓ (0.470mol / L) = <em>0.3525 moles of H₂SO₄</em>
KOH: 0.700L ₓ (0.240mol / L) = <em>0.168 moles of KOH</em>
The moles of sulfuric acis that react with KOH are:
0.168mol KOH ₓ (1 mole H₂SO₄ / 2 moles KOH) = 0.0840 moles of sulfuric acid.
Thus, moles that remain are:
0.3525moles - 0.0840 moles = <em>0.2685 moles of sulfuric acid remains</em>
As total volume is 0.700L + 0.750L = 1.450L, concentration is:
0.2685mol / 1.450L = <em>0.185M sulfuric acid</em>
