100 m = 0.1 km
9.58 sec = 9.58/3600 = 0.00266 hr
Speed = 0.1/0.00266= 37.6 km/hr
Can you mark it brainliest?
Answer And Explanation:
Option C is correct.
Lend and borrow money, taxation, law enforcement, charter banks and transportation.
Some of the powers that were mentioned in the other options that weren't concurrent powers (that is, they belong to either the state government alone or the federal government alone) & disqualified them from being the answer include:
National defence (federal), Currency (federal), foreign affairs (federal), intrastate commerce (state) etc.
This problem is to use the Claussius-Clapeyron Equation, which is:
ln [p2 / p1] = ΔH/R [1/T2 - 1/T1]
Where p2 and p1 and vapor pressure at estates 2 and 1
ΔH is the enthalpy of vaporization
R is the universal constant of gases = 8.314 J / mol*K
T2 and T1 are the temperatures at the estates 2 and 1.
The normal boiling point => 1 atm (the pressure of the atmosphere at sea level) = 101,325 kPa
Then p2 = 101.325 kPa
T2 = ?
p1 = 54.0 kPa
T1 = 57.8 °C + 273.15K = 330.95 K
ΔH = 33.05 kJ/mol = 33,050 J/mol
=> ln [101.325/54.0] = [ (33,050 J/mol) / (8.314 J/mol*K) ] * [1/x - 1/330.95]
=> 0.629349 = 3975.22 [1/x - 1/330.95] = > 1/x = 0.000157 + 1/330.95 = 0.003179
=> x = 314.6 K => 314.6 - 273.15 = 41.5°C
Answer: 41.5 °C
Answer:
π = 14.824 atm
Explanation:
wt % = ( w NaCL / w sea water ) * 100 = 3.5 %
assuming w sea water = 100 g = 0.1 Kg
⇒ w NaCl = 3.5 g
osmotic pressure ( π ):
∴ T = 20 °C + 273 = 293 K
∴ C ≡ mol/L
∴ density sea water = 1.03 Kg/L....from literature
⇒ volume sea water = 0.1 Kg * ( L / 1.03 Kg ) = 0.097 L sln
⇒ mol NaCl = 3.5 g NaCL * ( mol NaCL / 58.44 g ) = 0.06 mol
⇒ C NaCl = 0.06 mol / 0.097 L = 0.617 M
⇒ π = 0.617 mol/L * 0.082 atm L / K mol * 293 K
⇒ π = 14.824 atm
Answer:
6.68 X 10^-11
Explanation:
From the second Ka, you can calculate pKa = -log (Ka2) = 6.187
The pH at the second equivalence point (8.181) will be the average of pKa2 and pKa3. So,
8.181 = (6.187 + pKa3) / 2
Solving gives pKa3 = 10.175, and Ka3 = 10^-pKa3 = 6.68 X 10^-11
