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3 years ago

15

Which line segment on the graph represents boiling

1 answer:

hoa [83]3 years ago

8 0

1. There is no graph given

2. ENERGY LEVELS!!!

Boiling is the process of liquid(middle energy, flat line) turning into a gas(lots of energy, flat line), hope this helps

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Calculate the pH of the 4M NaOH

krek1111 [17]

$\text{NaOH}(aq) \longrightarrow \text{Na}^{+} (aq)+ \text{OH}^{-}(aq)\\\\k_w = [\text{H}_3 \text{O}^{+}][\text{OH}^{-}]\\\\\implies 10^{-14} = [\text{H}_3 \text{O}^{+}] \cdot 4\\\\\implies [\text{H}_3 \text{O}^{+}] = \dfrac{10^{-14}}4 = 2.5 \times 10^{-15}~ M\\\\\text{pH} = -\log[\text{H}_3 \text{O}^{+}] = -log(2.5 \times 10^{-15})=14.603$

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2 years ago

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For the reaction 2NH3(g) + 2O2(g)N2O(g) + 3H2O(l) H° = -683.1 kJ and S° = -365.6 J/K The standard free energy change for the rea

BlackZzzverrR [31]

Answer:

$\Delta G^{0}$ = -457.9 kJ and reaction is product favored.

Explanation:

The given reaction is associated with 2 moles of $NH_{3}$

Standard free energy change of the reaction ( $\Delta G^{0}$ ) is given as:

$\Delta G^{0}=\Delta H^{0}-T\Delta S^{0}$ , where T represents temperature in kelvin scale

So, $\Delta G^{0}=(-683.1\times 10^{3})J-(273K\times -365.6J/K)=-583291.2J$

So, for the reaction of 1.57 moles of $NH_{3}$ , $\Delta G^{0}=(\frac{1.57}{2})\times -583291.2J=-457883.592J=-457.9kJ$

As, $\Delta G^{0}$ is negative therefore reaction is product favored under standard condition.

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52. A proton is a subatomic particle that has a

kondaur [170]

Answer:

positive charge

Explanation:

Protons are positively charged

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The is the B) doing background research about the question

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Which term refers to the speed at which reactants are converted into products?

shutvik [7]

Answer:

activation energy hope this is right

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