Answer:
Increase the pressure of the gas
Explanation:
According to the Pressure law, for a fixed mass of gas, at a constant volume (V), the pressure (P) is directly proportional to the absolute temperature (T).
From the kinetic molecular theory, gases are composed of particles which are in constant motion, colliding with themselves as well as with the walls of their container.
When the temperature of these gas molecules is increased, the molecules acquire more kinetic energy and the rate of collisions increases. Since the container cannot expand, the increase in pressure is due to the increase in collisions between the molecules of the gas as well as with the walls of their container.
Answer:
D 1 and 3 only I am not sure
Explanation:
A COVALENT BOND, FORMS BETWEEN ELEMENTS WITH SIMILAR ELECTRONEGATIVITY AS SHARING OF ELECTRON PAIRS BETWEEN ATOMS IS EASIER AS THEY ARE IDENTICAL.
Explanation:
Bonding atoms with similar electronegativity values form covalent bonds.
A covalent bond, also called a molecular bond, is a chemical bond that involves the sharing of electron pairs between atoms.
Covalent bonds form between two nonmetal atoms with identical or relatively close electronegativity values
Electronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons, also it is the strength an atom has to attract a bonding pair of electrons to itself.
Pure covalent bonds result when two atoms of the same electronegativity bond. This occurs only when two atoms of the same element bond with each other.
Answer:
-1
Explanation:
The relation between Kp and Kc is given below:
Where,
Kp is the pressure equilibrium constant
Kc is the molar equilibrium constant
R is gas constant
, 0.082057 L atm.mol⁻¹K⁻¹
T is the temperature in Kelvins
Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)
For the first equilibrium reaction:
<u>Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants) = (2+1)-(2+2) = -1 </u>
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Answer
is: 0.375 moles are present in 8.4 liters of nitrous oxide at stp.
V(N₂O) = 8.4 L.
V(N₂O) =
n(N₂O) · Vm.
Vm = 22,4 L/mol.<span>
n</span>(N₂O) = V(N₂O) ÷ Vm.
n(N₂O) = 8.4 L ÷ 22.4 L/mol.
n(N₂O) = 0.375 mol.<span>
Vm - molare volume on STP.</span>
