Answer:
At STP, 760mmHg or 1 atm and OK or 273 degrees celcius
Explanation:
The standard temperature and pressure is the temperature and pressure at which we have the molecules of a gas behaving as an ideal gas. At this temperature and pressure, it is expected that the gas exhibits some properties that make it behave like an ideal gas.
This temperature and pressure conform some certain properties on a gas molecule which make us say it is behaving like an ideal gas. Ordinarily at other temperatures and pressures, these properties are not obtainable
Take for instance, one mole of a gas at stp occupies a volume of 22.4L. This particular volume is not obtainable at other temperatures and pressures but at this particular temperature and pressure. One mole of a gas will occupy this said volume no matter its molar mass and constituent elements. This is because at this temperature and pressure, the gas is expected to behave like an ideal gas and thus exhibit the characteristics which are expected of an ideal gas
Answer:
I HOPE THE ABOVE INFORMATION WILL HELP YOU A LOT.
HAVE A NICE DAY.
Answer:
Acceleration = (change in speed) / (time for the change)
Change in speed= (0 - 26 km/hr) = -26 km/hr
(-26 km/hr) x (1,000 m/km) x (1 hr / 3,600 sec) = -7.222 m/sec
Average acceleration = (-7.222 m/s) / (22 min x 60sec/min) = -0.00547 m/sec²
Average speed during the stopping maneuver =
(1/2) (start speed + end speed) = 13 km/hr = 3.6111 m/sec
Explanation:
Answer:
E₁ ≅ 28.96 kJ/mol
Explanation:
Given that:
The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol,
Let the activation energy for a catalyzed biochemical reaction = E₁
E₁ = ??? (unknown)
Let the activation energy for an uncatalyzed biochemical reaction = E₂
E₂ = 50.0 kJ/mol
= 50,000 J/mol
Temperature (T) = 37°C
= (37+273.15)K
= 310.15K
Rate constant (R) = 8.314 J/mol/k
Also, let the constant rate for the catalyzed biochemical reaction = K₁
let the constant rate for the uncatalyzed biochemical reaction = K₂
If the rate constant for the reaction increases by a factor of 3.50 × 10³ as compared with the uncatalyzed reaction, That implies that:
K₁ = 3.50 × 10³
K₂ = 1
Now, to calculate the activation energy for the catalyzed reaction going by the following above parameter;
we can use the formula for Arrhenius equation;
If 
&
E₁ ≅ 28.96 kJ/mol
∴ the activation energy for a catalyzed biochemical reaction (E₁) = 28.96 kJ/mol
