Answer: When the pressure exerted on a confined gas at constant temperature is doubled, the volume of the gas is halved.
Explanation:
To calculate the new volume, we use the equation given by Boyle's law. This law states that pressure is inversely proportional to the volume of the gas at constant temperature.
The equation given by this law is:
where,
= initial pressure of gas = p
= initial volume of gas = v
= final pressure of gas= 2p
= final volume of gas = ?
Putting values in above equation, we get:
Thus the final volume of the gas is half of the initial volume.
(2) 200. K to 400. K (4) 200.°C to 400.°C
What volume of oxygen at STP is required for the complete combustion of 100.50 mL of C2H2?
201 mL
201.00 mL
251 mL
251.25 mL
251.25 mL of O₂
The balance chemical equation is as follow,
2 C₂H₂ + 5 O₂ → 4 CO₂ + 2 H₂O
As we know if the gas is acting ideally then 1 mole of any gas at standard temperature and pressure it will occupy exactly 22.4 L or 22400 mL of volume.
Keeping this in mind according to equation,
44800 mL (2 mol) of C₂H₂ required = 112000 mL (5 mol) of O₂
So,
100.50 mL of C₂H₂ will require = X mL of O₂
Solving for X,
X = (100.5 mL × 112000 mL) ÷ 44800 mL
X = 251.25 mL of O₂
The balanced reaction would be:
C2H2 + 5/2O2 = 2CO2 + H2O
We are given the amount of acetylene in the reaction. This will be the starting point of our calculation. We use the ideal gas equation to find for the number of moles.
n = PV / RT = 1.00(.1005 L) / (0.08206 atm L/mol K ) 273.15 K
n= 4.4837 x 10^-3 mol C2H2
4.4837 x 10^-3 mol C2H2 (5/2 mol O2/ 1 mol C2H2) = 0.0112 mol O2
V = nRT/P = 0.0112 mol O2 x 273.15 K x 0.08206 atm L/mol K / 1 atm
V=0.25125 L or 251.25 mL
b. suspensions
c. colloids
d. solutions
The answer is A) heterogeneous mixtures, look up how a heterogeneous mixture is separated and it will tell you distillation. Have a nice day and good luck on the rest of your work
(2) low first ionization energy and high electronegativity
(3) high first ionization energy and low electronegativity
(4) high first ionization energy and high electronegativity
Answer:
low first ionization energy and low electronegativity
Explanation:
Metals are typically characterized by low first ionization energy and low electronegativity. This is because metals can easily lose electrons to become positive ions and do not attract electrons strongly.
Two properties that are typically associated with metals are low first ionization energy and low electronegativity. Therefore, the correct option is (1).
Ionization energy is the energy required to remove an electron from an atom. Metals have low first ionization energy because they can easily lose an electron to become positive ions. Electronegativity is a measure of the ability of an atom to attract electrons. Metals have low electronegativity because they do not attract electrons strongly.
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