In dilution we add distilled water to decrease the concentration of required sample from high concentration to lower concentration
The law used for dilution:
M₁V₁]Before dilution = M₂V₂] After dilution
M₁ = 1.5 M
V₁ = ?
M₂ = 0.3 M
V₂ = 500 ml
1.5 * V₁ = 0.3 * 500 ml
so V₁ = 100 ml and it completed to 500 ml using 400 ml deionized water
Answer:
a) Pabs = 48960 KPa
b) T = 433.332 °C
Explanation:
∴ d = 1000 Kg/m³
∴ g = 9.8 m/s²
∴ h = 5000 m
∴ P gauge = - 40 KPa * ( 1000 Pa / KPa ) = - 40000 Pa; Pa≡Kg/m*s²
⇒ Pabs = - 40000 Kg/ms² + ( 1000 Kg/m³ * 9.8 m/s² * 5000 m )
⇒ Pabs = 48960000 Pa = 48960 KPa
a) at that height and pressure, we find the temperature at which the water boils by means of an almost-exponential graph which has the following equation:
P(T) = 0.61094 exp ( 17.625*T / ( T + 243.04 ))......P (KPa) ∧ T (°C)....from literature
∴ P = 48960 KPa
⇒ ( 48960 KPa / 0.61094 ) = exp ( 17.625T / (T+ 243.04))
⇒ 80138.803 = exp ( 17.625T / ( T + 243.04))
⇒ Ln ( 80138.803) = 17.625T / ( T + 243.04))
⇒ 11.292 * ( T + 243.04 ) = 17.625T
⇒ 11.292T + 2744.289 = 17.625T
⇒ 2744.289 = 17.625T - 11.292T
⇒ 2744.289 = 6.333T
⇒ T = 433.332 °C
Right answer is B . Trust me .
