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.
<h2>Answer:</h2>
He is right that the energy of vaporization of 47 g of water s 106222 j.
<h3>Explanation:</h3>
Enthalpy of vaporization or heat of vaporization is the amount of energy which is used to transform one mole of liquid into gas.
In case of water it is 40.65 KJ/mol. And 18 g of water is equal to one mole.
It means for vaporizing 18 g, 40.65 kJ energy is needed.
So for energy 47 g of water = 47/18 * 40.65 = 106.1 KJ
Hence the student is right about the energy of vaporization of 47 g of water.
Answer:
2 H2(g) + O2(g) → 2 H2O(ℓ) ΔH = −570 kJ
Explanation:
Well, you could get the mass as
ad then
, where
is the sea level weight,
the sea level accel.,
the accel. above while
the weight above.
