Answer:
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Answer:
From point, 1 mole of water = molar mass of water =18g 20 moles of water = 18 g x 20 = 360g (iv) From point, 6.022 x 1023 molecules of water = 1 mole = 18g of water 1.2044 x 1025 molecules of water Therefore, points (ii) and (iv) represent 360 g of water.
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
Answer:
2.00 moles of Ni has 1.2 *10^24 atoms
Explanation:
Step 1: Data given
Number of moles Ni = 2.00 moles
Number of Avogadro = 6.022*10^23 /mol
Step 2: Calculate number of atoms
Number of particles (=atoms) = Number of Avogadro * number of moles
Number of atoms = 6.022 * 10^23 /mol * 2.00 moles
Number of atoms = 1.2*10^24 atoms
2.00 moles of Ni has 1.2 *10^24 atoms
Answer:
A , B, C
Explanation: D is a Diamagnetic
