Answer:
C. It decreases by a factor of 4
Explanation:
F1 = kq1*q2/r²
F2 = kq1*q2/(2r)² = kq1*q2/(4r²) = kq1*q2/(r²*4) = F1/4
A sample of a compound contains 60.0 g C and 5.05 g H.
divide by molar mass of C(12) and H(1) to get molar ratio
C: 60/12=5 and H: 5/1=5
so C:H=5:5=1:1
total molar mass=78
divide by 1C+1H to find the formula: 78/(12+1)=78/13=6
compound is C6H6
The reaction;
O(g) +O2(g)→O3(g), ΔH = sum of bond enthalpy of reactants-sum of food enthalpy of products.
ΔH = ( bond enthalpy of O(g)+bond enthalpy of O2 (g) - bond enthalpy of O3(g)
-107.2 kJ/mol = O+487.7kJ/mol =O+487.7 kJ/mol +487.7kJ/mol =594.9 kJ/mol
Bond enthalpy (BE) of O3(g) is equals to 2× bond enthalpy of O3(g) because, O3(g) has two types of bonds from its lewis structure (0-0=0).
∴2BE of O3(g) = 594.9kJ/mol
Average bond enthalpy = 594.9kJ/mol/2
=297.45kJ/mol
∴ Averange bond enthalpy of O3(g) is 297.45kJ/mol.
Length of 1 side 1.2*10^-5km =1.2*10^-5*10^5 =1.2cm
<span>volume of the cube (1.2)^3=1.728 cm^3 </span>
<span>density= mass/volume= 1.1/1.728=0.636 g/cm^3</span>
At higher temperature, and lower pressure.
