Rock paper scissor and that is on Albert Einstein , he made that up because he is so freaking smart so there you go have a wonderful day
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
According to this formula :
㏑[A] /[Ao] = - Kt
when we have Ao = 0.3 m
and K =0.46 s^-1
t = 20min = 0.2 x 60 =12 s
So by substitution :
㏑[A] / 0.3 = - 0.46 * 12
㏑[A] / 0.3 = - 5.52
by taking e^x for both side of the equation we can get [A]
∴[A] = 0.0012 mol dm^-3
Answer:
3853 g
Step-by-step explanation:
M_r: 107.87
16Ag + S₈ ⟶ 8Ag₂S; ΔH°f = -31.8 kJ·mol⁻¹
1. Calculate the moles of Ag₂S
Moles of Ag₂S = 567.9 kJ × 1 mol Ag₂S/31.8kJ = 17.858 mol Ag₂S
2. Calculate the moles of Ag
Moles of Ag = 17.86 mol Ag₂S × (16 mol Ag/8 mol Ag₂S) = 35.717 mol Ag
3. Calculate the mass of Ag
Mass of g = 35.717 mol Ag × (107.87 g Ag/1 mol Ag) = 3853 g Ag
You must react 3853 g of Ag to produce 567.9 kJ of heat
Answer:the force from spinning that moves things away from the center
Explanation: ap chem
