<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.
Given the molar mass of Nitrogen is 14.01g/mol you can use that to solve for the moles of nitrogen.
0.235g(1mol/14.01g) = .0168 moles.
Ca(s)+2Hcl(aq) ------>CaCl2(s)+H2(g)
1) The forward reaction is N2 (g) + O2 (g) → 2NO
(that reaction requires special contitions because at normal pressures and temperatures N2 and O2 do not react to form another compound.
2) The equiblibrium equation is
N2 (g) + O2 (g) ⇄ 2NO
3) Then, the reverse reaction is
2NO → N2(g) + O2(g)
Answer: 2NO → N2(g) + O2(g)
Answer:
23.8g of sodium phosphate are formed
Explanation:
Based on the reaction of sodium, Na, with phosphoric acid, H₃PO₄:
3Na + H₃PO₄ → Na₃PO₄ + 3/2 H₂
<em>3 moles of sodium produce 1 mole of sodium phosphate</em>
<em />
To solve this question we must find the moles of sodium in 10g. Using the chemical reaction we can find the moles -And the mass- of sodium phosphate produced, as follows:
<em>Moles Na -Molar mass: 22.99g/mol-</em>
10g * (1mol / 22.99g) = 0.435 moles Na
<em>Moles Na₃PO₄:</em>
0.435 moles Na * (1mol Na₃PO₄ / 3mol Na) = 0.145 moles Na₃PO₄
<em>Mass Na₃PO₄ -Molar mass: 163.94g/mol-</em>
0.145 moles Na₃PO₄ * (163.94g/mol) =
<h3>23.8g of sodium phosphate are formed</h3>
