Thomson saw the atom to be a spherical cloud of positive proton matter with electrons dispersed throughout it. Then the gold foil experiment is when Rutherford shot alpha particles at a sheet of gold foil and when there was deflection he concluded there was a dense part in the center of an atom with he called the nucleus made up of neutrons and protons.
C(HClO) = 0,3 M.
<span>V(HClO) = 200 mL = 0,2 L.
n(HClO) = </span>c(HClO) · V(HClO).
n(HClO) = 0,06 mol.<span>
c(KClO</span>) =
0,2 M.
<span>V(KClO) = 0,3 L.
n(KClO) = 0,06 mol.
V(buffer solution) = 0,2 L + 0,3 L = 0,5 L.
ck</span>(HClO) = 0,06 mol ÷ 0,5 L = 0,12 M.
cs(KClO) = 0,06 mol ÷ 0,5 L = 0,12 M.<span>
Ka(HClO</span>) =
2,9·10⁻⁸.<span>
This is buffer solution, so use Henderson–Hasselbalch
equation:
pH = pKa + log(cs</span> ÷ ck).<span>
pH = -log(</span>2,9·10⁻⁸) + log(0,12 M ÷ 0,12 M).<span>
pH = 7,54 + 0.
pH = 7,54</span>
It’s c or b I think I’m not sure
The value of ΔH° when 1.00 gal of gasoline burns to produce carbon dioxide gas and water vapour is -118. 167kJ.
<h3>Chemical reaction:</h3>
C8H18 + 25/2 O2 ----- 8CO2 +9H2O
<h3>Calculation of heat of reaction</h3>
∆H = ∆H°(product) - ∆H°(reactant)
∆H = ∆H° (CO2) +∆H°(H2O) - ∆H°(C8H18) -∆H° (O2)
As we know that,
∆H° (CO2) =-393.5 kJ/mol
∆H°(H2O) = - 241.836 kJ/mol
∆H°(C8H18) = -250.1 kJ/mol
∆H°(O2) = 0 kJ/mol
By substituting all the values, we get
∆H = 8(-393.5 kJ/mol) + 9(- 241.836 kJ/mol) +250.1 + 0
∆H = -5074.33 kJ/mol
<h3>Calculation of heat released when 1 gal of gasoline is burned</h3>
This can be given as
= (-5074.334 × 0.7028 × 1000 × 3.785)/114.23
By calculating we obtained
∆H° = -118.167kJ.
Thus, we calculated that the value of ΔH° when 1.00 gal of gasoline burns to produce carbon dioxide gas and water vapour is -118. 167kJ.
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Explanation:
how are you confused there just tell me the problem