Answer:
C. Carbon
Explanation:
Carbon has an electronegativity of 2.55, followed by Tin at 1.96, Silicon at 1.90 and the least electronegative would be Lead at 1.87.
Answer:
b. CH₂Cl₂ is more volatile than CH₂Br₂ because of the large dispersion forces in CH₂Br₂
Explanation:
CH₂Cl₂ is more volatile than CH₂Br₂ (b.p of CH₂Cl₂ = 39,6 °C; b.p of CH₂Br₂ = 96,95°C). Thus, c. and d. are FALSE
Dipole-dipole interactions in CH₂Cl₂ are greater than the dipole-dipole interactions in CH₂Br₂ because Cl is more electronegative that Br (Cl = 3,16; Br = 2,96). But this mean CH₂Cl₂ is less volatile than CH₂Br₂ but it is false.
There are large dispersion forces in CH₂Br₂ because Br has more electrons and protons than Cl. Large disperson forces mean CH₂Br₂ is less volatile than CH₂Cl₂ and it is true.
I hope it helps!
Compounds are classified according to the elements that make them up. For example, oxides contain one or more oxygen atoms, hydrides contain one or more hydrogen atoms.
compounds form different types of bonds too. a metal and nonmetal element will create an ionic bond, two nonmetal elements create covalent bonds
Answer:
the entropy change for the surroundings when 1.62 moles of CH4(g) react at standard conditions is −8.343 J/K
Explanation:
The balanced chemical equation of the reaction in the question given is:
Using standard thermodynamic data at 298K.
The entropy of each compound above are listed as follows in a respective order.
Entropy of (CH4(g)) = 186.264 J/mol.K
Entropy of (O2(g)) = 205.138 J/mol.K
Entropy of (CO2(g)) = 213.74 J/mol.K
Entropy of (H2O(g)) = 188.825 J/mol.K
The change in Entropy (S) of the reaction is therefore calculated as follows:
= -5.15 J/mol.K
Given that :
the number of moles = 1.62 of CH4(g) react at standard conditions.
Then;
The change in entropy of the rxn 
= −8.343 J/K
To find the neutrons you add the mass number and the atomic number for neutrons
Hope this helps
-Zayn Malik
