Answer:
1) When 6.97 grams of sodium(s) react with excess water(l), 56.0 kJ of energy are evolved.
2) When 10.4 grams of carbon monoxide(g) react with excess water(l), 1.04 kJ of energy are absorbed.
Explanation:
1) The following thermochemical equation is for the reaction of sodium(s) with water(l) to form sodium hydroxide(aq) and hydrogen(g).
2 Na(s) + 2H₂O(l) ⇒ 2NaOH(aq) + H₂(g) ΔH = -369 kJ
The enthalpy of the reaction is negative, which means that 369 kJ of energy are evolved per 2 moles of sodium. The energy evolved for 6.97 g of Na (molar mass 22.98 g/mol) is:
2) The following thermochemical equation is for the reaction of carbon monoxide(g) with water(l) to form carbon dioxide(g) and hydrogen(g).
CO(g) + H₂O(l) ⇒ CO₂(g) + H₂(g) ΔH = 2.80 kJ
The enthalpy of the reaction is positive, which means that 2.80 kJ of energy are absorbed per mole of carbon monoxide. The energy evolved for 10.4 g of CO (molar mass 28.01 g/mol) is:
Answer:
<h2>HCL ( Hydrogen-Chloride )</h2>
The larger the paper airplane the more it will weigh, the more it weighs the more lift will be needed to keep it flying. Eventually weight will become greater than lift and the paper airplane will decend to the ground. In adition the larger the paper airplane the larger its wings can be.
Answer: A. 0.0375
Explanation:
To calculate the moles :
According to stoichiometry :
As 5 moles of 
give = 2 moles of 
Thus 0.094 moles of give =
of
Thus 0.0375 moles of will be produced when 3.00g of O₂ react completely
Answer:
[Ar] 3d10 4s1
Explanation:
The correct electronic configuration of copper is [Ar] 3d10 4s1
Copper has atomic number 29 and due to the stability of half filled or fully filled orbitals or shells, the electrons from the 4s jumps to the 3d and makes the 3d shell contain 10 electrons.
This is what I mean:
Cu = Ar 4s2 3d10 is the expected configuration of copper when we follow the principle of filling the various orbitals that is the s, p, d, f orbitals.
But because we write 3d before writing 4s, we have Ar 3d10 4s2. Instead of this configuration becoming the correct one, an electron from the 4s orbital jumps to the 3d orbital to complete the orbital giving the electrons in the 3d orbital 10.
So therefore the correct configuration is Ar 3d10 4s1
