Answer:
0.00915 M of
remain after 5.16 seconds.
Explanation:
Using integrated rate law for first order kinetics as:
Where,
is the concentration at time t
is the initial concentration
Given that:
The rate constant, k =
s⁻¹
Initial concentration
= 0.054 M
Final concentration
= ? M
Time = 5.16 s
Applying in the above equation, we get that:-
<u>0.00915 M of
remain after 5.16 seconds.</u>
The sample of oxygen gas was collected through water displacement. So, the gas collected will be a mixture of oxygen and water vapor.
Given that the total pressure of the mixture of gases containing oxygen and water vapor = 749 Torr
Vapor pressure of pure water at
=25.81mmHg
= 
According to Dalton's law of partial pressures,
Total pressure = Partial pressure of Oxygen gas + Partial pressure of water
749 Torr = Partial pressure of Oxygen gas + 25.81 Torr
Partial pressure of Oxygen gas = 749 Torr - 25.81 Torr = 723.19 Torr
Therefore the partial pressure of Oxygen gas in the mixture collected will be 723.19 Torr
Explanation:
Mixture is the physical Combination Of two or Substance
Example
a mixture of sugar and water.
Compound is the chemical combination of two or more metals.
Example.
a mixture of hydrogen and water.
A mixture of hydrogen and oxygen forms water or H2O
whereas The dihydrogen monoxide parody involves calling water by an unfamiliar chemical name, most often "dihydrogen monoxide" (DHMO), and listing some of water's properties in a particularly alarming manner, such as accelerating corrosion (rust) and causing suffocation (drowning). The parody often calls for dihydrogen monoxide to be banned, regulated, or labeled as dangerous. It plays into chemophobia and demonstrates how a lack of scientific literacy and an exaggerated analysis can lead to misplaced fears. The parody has been used with other chemical names such as hydrogen hydroxide, dihydrogen oxide, hydroxic acid, hydric acid and oxidane.
Answer:
0.18 moles
Explanation:
Applying,
PV = nRT................... Equation 1
Where P = pressure, V = volume, n = number of moles, R = molar gas constant, T = temperature.
make n the subject of the equation
n = PV/RT............... Equation 2
Given: V = 5.3 L, T = 22 °C = (22+272) K = 295 K, P = 632 mmHg = (0.00131579×632) = 0.8316 atm, R = 0.083 L.atm/K.mol
Substitute these values into equation 2
n = (0.8316×5.3)/(0.083×295)
n = 0.18 moles