Hydrogen bonds.
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Newton's third law states that every action has an equal and opposite reaction. The action and reaction forces are pairs of opposing forces.
In the given examples all three obey Newton's third law.
B) Action force: John pulls the door handle
Reaction: door handle gets pulled
C) Action force: Tire pushes on road
Reaction: The road pushes on the tire, vehicle moves
D) Action force: Exhaust pushes out of a rocket
Reaction: Rocket is pushed forward
Ans A) All these are examples of Newtons third law
Answer:
1) 0 C2H4O2 + 0 O2 -> 0 CO2 + 0 H2O (balanced)
2) V2O5 + CaS -> CaO + V2S5
<em>just additional info: V2O5 </em><em>is</em><em> </em><em>divanadium</em><em> </em><em>pentaoxide</em>
LHS (Left hand side)
V: 2
O: 5
Ca: 1
S: 1 x 5 [to balance with the right hand side of the equation]
RHS (Right hand side)
V: 2
O: 1 x 5 [to balance with the left hand side of the equation]
Ca: 1
S: 5
When you balance any elements, you have to balance the whole chemical compound.
Thus,
V2O5 + <em><u>5</u></em> CaS -> <em><u>5</u> CaO</em> + V2S5
LHS CHECK:
V: 2
O: 5
Ca: 5
S: 5
RHS CHECK:
V: 2
O: 5
Ca: 5
S: 5
3) S8 + O2 -> SO2
LHS:
S: 8
O: 2
RHS:
S: 1 x 8 [to balance with LHS]
O: 2
When you balance any elements, you have to balance the whole chemical compound.
S8 + O2 -> <em><u>8</u></em><em><u> </u></em>SO2
When we add 8 to the RHS, it gives us 8S, 16 O.
In order to balance that into the RHS, I need to multiply the O2 by 8, which will give 8(O2) = 16 O particles.
Therefore, <em><u>S8</u></em><em><u> </u></em><em><u>+</u></em><em><u> </u></em><em><u>8</u></em><em><u> </u></em><em><u>O2</u></em><em><u> </u></em><em><u>-</u></em><em><u>></u></em><em><u> </u></em><em><u>8</u></em><em><u> </u></em><em><u>SO2</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>final</u></em><em><u> </u></em><em><u>answer</u></em><em><u> </u></em><em><u>for</u></em><em><u> </u></em><em><u>(</u></em><em><u>3</u></em><em><u>)</u></em><em><u>.</u></em>
Answer:
Change in entropy for the reaction is
ΔS° = -268.13 J/K.mol
Explanation:
To calculate the change in entropy for the balanced reaction, we require the natural entropy of all the reactants and products in the reaction.
3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)
From Literature.
S°(NO₂) = 240.06 J/K.mol
S°(H₂O) = 69.91 J/K.mol
S°(HNO₃) = 155.60 J/K.mol
S°(NO) = 210.76 J/K.mol
These are the entropies of the reactants and products under standard conditions of 298.15 K and 1 atm.
Note that
ΔS° = Σ nᵢS°(for products) - Σ nᵢS°(for reactants)
Σ nᵢS°(for products) = [2 × S°(HNO₃)] + [1 × S°(NO)]
= (2 × 155.60) + (1 × 210.76) = 521.96 J/K.mol
Σ nᵢS°(for reactants) = [3 × S°(NO₂)] + [1 × S°(H₂O)]
= (3 × 240.06) + (1 × 69.91) =790.09 J/K.mol
ΔS° = Σ nᵢS°(for products) - Σ nᵢS°(for reactants)
ΔS° = 521.96 - 790.09 = -268.13 J/K.mol
Hope this Helps!!
Answer:
Explanation:
Mass box A = 10 grams; Mass box B = 5 grams; Mass box C—made of one A and one B How many grams of A would be needed to give the same number of particles: as 7.5 grams of B? 20 as 2.5 grams? 12.5 as 12.5 grams? 37.5 as 1 gram? 2 Count the number of molecules in the diagram of water formation given in the lesson to complete the following ...
