B. The partial pressure of N2 is 101 kPa
<h3>Further explanation</h3>
Given
volume = 22.4 L
1.0 mol of nitrogen and 2.0 mol of hydrogen at 0°C
Required
Total pressure and partial pressure
Solution
Ideal gas law :
PV = nRT
n total = 3 mol
T = O °C + 273 = 273 K
P = nRT/V
P = 3 x 0.08205 x 273 / 22.4
P total = 3 atm = 303,975 kPa
P Nitrogen = 1/3 x 303.975 = 101.325 kPa
P Hydrogen = 2/3 x 303.975 = 202.65 kPa
Isotones are nuclides that have the same neutron number but a different proton number. Therefore the answer is C. boron and carbon.
There are ALOT because they would always come in and out and they will burst which creates more so techneclly there are infinate
Answer:
4) Van der waals forces
Explanation:
Krypton (Kr) belongs to the noble gas group and has fully filled valence orbitals. In the solid phase, Kr exists as a white solid with a face centered cubic structure.
Intermolecular forces of attraction from the strongest to the weakest include:
Ionic > hydrogen bonding > dipole-dipole > london dispersion
Kr is monoatomic and non-polar. When fully filled (stable) valence orbitals of 2 Kr atoms approach each other in close proximity they experience a repulsive force which prevents the formation of strong bonds. Thus, the only force of attraction in Kr is the long range weak Van Der Waals force also known as the london dispersion force.
Answer:
e. 3.08 x 10⁻² mol of ions.
Explanation:
- Every 1.0 mole of any compound contains Avogadro's number of molecules (6.022 x 10²³).
- We can get the no. of moles of NiCl₂ using cross multiplication:
1.0 mol NiCl₂ contains → 6.022 x 10²³ molecules.
??? mol NiCl₂ contains → 6.188 x 10²¹ molecules.
∴ The no. of moles of NiCl₂ = (1.0 mol)(6.188 x 10²¹ molecules)/(6.022 x 10²³ molecules) = 1.028 x 10⁻² mol.
- NiCl₂ is ionized according to the equation:
NiCl₂ → Ni²⁺ + 2Cl⁻.
Which means that every 1.0 mol of NiCl₂ is ionized to produce 3.0 moles (1.0 mol of Ni²⁺ and 2 moles of Cl⁻).
<em>∴ The total moles of ions are released</em> = 3 x 1.028 x 10⁻² mol = <em>3.083 x 10⁻² mol of ions.</em>
