Answer:
A piece of gold foil was hit with alpha particles, which have a positive charge. Most alpha particles went right through. This showed that the gold atoms were mostly empty space. Some particles had their paths bent at large angles. A few even bounced backward. The only way this would happen was if the atom had a small, heavy region of positive charge inside it.
Answer:
-85 °C
Explanation:
O and S are in the same group( Group 16). Since S is below O it's atomic mass is higher than O. So molar mass of H2S is higher than H2O. The strength of Vanderwaal Interactions ( London dispersion forces) increases when the molar mass increases. However, only H2O can form H bonds with each other. This is because electronegativity of O is higher than S and therefore H in H2O has a higher partial positive charge than H of H2S.
H bond dominate among these 2 types of forces so the strength of attractions between molecules is higher in H2O than H2S. Therefore more energy should be supplied for H2O to break inter
molecular forces and convert from solid to liquid state than H2S. So mpt of H2O must be higher than that of H2S.
Answer:
ΔH of the reaction is -802.3kJ.
Explanation:
Using Hess's law, you can know ΔH of reaction by the sum of ΔH's of half-reactions.
Using the reactions:
<em>(1) </em>Cgraphite(s)+ 2H₂(g) → CH₄(g) ΔH₁ = −74.80kJ
<em>(2) </em>Cgraphite(s)+ O₂(g) → CO₂(g) ΔH₂ = −393.5k
J
<em>(3) </em>H₂(g) + 1/2 O₂(g) → H₂O(g) ΔH₃ = −241.80kJ
The sum of (2) - (1) produce:
CH₄(g) + O₂(g) → CO₂(g) + 2H₂(g) ΔH' = -393.5kJ - (-74.80kJ) = -318.7kJ
And the sum of this reaction with 2×(3) produce:
CH₄(g) + 2 O₂(g) → CO₂(g) + 2H₂O(g) And ΔH = -318.7kJ + 2×(-241.80kJ) =
<em>-802.3kJ</em>
