Answer:
Work done, W = 100 J
Explanation:
We have, Billy the friendly Robot uses 50 N of force to lift a box 2 meters in the air.
It is required to find the work done by Billy.
Work done by an object is given in terms of force and displacement. The formula used to find the work done is given by :

So, the work performed by Billy is 100 J.
<span>1.8 × (2/3) = 1.2
brainliest please
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Answer: Option (B) is the correct answer.
Explanation:
A covalent compound is a compound formed by sharing of electrons. And, in a covalent network solid atoms are bonded by covalent bonds in a continuous network that is extending throughout the material or solid.
This continuous arrangement of atoms are like a lattice.
For example, diamond is a covalent network solid in which carbon atoms are arranged in a continuous lattice like structure.
Hence, we can conclude that the statement all the atoms are covalently bonded to other atoms to form a lattice-like structure, best describes the structure of covalent network solids.
Answer: The given pairs of electrons most likely reside in
type of orbital.
Explanation:
As it is given that two lone pair of electrons are present on the oxygen atom of ketone (such as cyclohexanone).
Also, there will be one bond pair between carbon and oxygen atom.
Hence, total electrons present in the domain are as follows.
2 lone pairs + 1 bon pair of electron = 3 electron domains
This means that there will be
type of orbital present.
Thus, we can conclude that given pairs of electrons most likely reside in
type of orbital.
Answer:
-88.66 kJ/mol
Explanation:
The expressions of heat capacity (Cp,m) for C(s) and for H₂(g) are:
C(s): Cp,m/(J K-1 mol-1) = 16.86 + (4.77T/10³) - (8.54x10⁵/T²)
H₂(g): Cp,m/(J K-1 mol-1) = 27.28 + (3.26T/10³) + (0.50x10⁵/T²)
Cp = A + BT + CT⁻²
For the Kirchoff's Law:
ΔHf = ΔH°f + 
Where ΔH°f is the enthalpy at 298 K, T1 is 298 K, T2 is the temperature given (373 K), and DCp is the variation of Cp (products less reactants). ΔH°f for ethene is -84.68 kJ/mol and the reaction is:
2C(s) + 3H₂(g) → C₂H₆
So, DCp:
dA = A(C₂H₆) - [2xA(C) + 3xA(H₂)] = 14.73 - [2x16.86 + 3x27.28] = -100.83
dB = B(C₂H₆) - [2xB(C) + 3xB(H₂)] = 0.1272 - [2x4.77x10⁻³ + 3x3.26x10⁻³] = 0.10788
dC = C(C₂H₆) - [2xC(C) + 3xC(H₂)] = 0 - (2x(-8.54x10⁵) + 3x0.50x10⁵) = 15.58x10⁵
dCp = -100.83 + 0.10788T + 15.58x10⁵T⁻²
= -3796.48 J/mol = -3.80 kJ/mol (solved by a graphic calculator)
ΔHf = -84.68 - 3.80
ΔHf = -88.66 kJ/mol