Explanation:
Lithium is an electropositive element that readily loses electrons.
Oxygen is electronegative and it will readily accept electrons.
Due to this significant electronegativity differences between the two species they form electrovalent or ionic bonds between them.
2atoms of Li lose two electrons:
Li → Li²⁺ + e⁻
Lithium isoelectronic with helium
For oxygen;
O + 2e⁻ → O²⁻
Oxygen is isoelectronic with Neon
Two ions of the lithium combines with the oxygen to form the bond;
4Li + O₂ → 2Li₂O
The electrostatic attraction between the two ions forms the ionic bond
Answer:
d) cut the large sized Cu solid into smaller sized pieces
Explanation:
The aim of the question is to select the right condition for that would increases the rate of the reaction.
a) use a large sized piece of the solid Cu
This option is wrong. Reducing the surface area decreases the reaction rate.
b) lower the initial temperature below 25 °C for the liquid reactant, HNO3
Hugher temperatures leads to faster reactions hence this option is wrong.
c) use a 0.5 M HNO3 instead of 2.0 M HNO3
Higher concentration leads to increased rate of reaction. Hence this option is wrong.
d) cut the large sized Cu solid into smaller sized pieces
This leads to an increased surface area of the reactants, which leads to an increased rate of the reaction. This is the correct option.
Hey there!
The answer as well as the explanation is in the image attached. Let me know if there's anything you're unable to see.
Hope this helps!
Answer:
The answer is SiO2
Explanation:
Silocon dioxide is written without a 1 after the silocon and with a 2 after the oxygen.
Answer:
fH = - 3,255.7 kJ/mol
Explanation:
Because the bomb calorimeter is adiabatic (q =0), there'is no heat inside or outside it, so the heat flow from the combustion plus the heat flow of the system (bomb, water, and the contents) must be 0.
Qsystem + Qcombustion = 0
Qsystem = heat capacity*ΔT
10000*(25.000 - 20.826) + Qc = 0
Qcombustion = - 41,740 J = - 41.74 kJ
So, the enthaply of formation of benzene (fH) at 298.15 K (25.000 ºC) is the heat of the combustion, divided by the number of moles of it. The molar mass od benzene is: 6x12 g/mol of C + 6x1 g/mol of H = 78 g/mol, and:
n = mass/molar mass = 1/ 78
n = 0.01282 mol
fH = -41.74/0.01282
fH = - 3,255.7 kJ/mol
