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Helga [31]
3 years ago
14

With regard to wind, describe the time of day that an early explorer might have planned to enter a harbor and when he might have

planned to leave for his trip home. Explain the reasoning by utilizing the concept of local breezes.
Chemistry
2 answers:
Alinara [238K]3 years ago
5 0

Early explorers planned to enter a harbor the sea during high tide or during tide rise.

It will be better to enter with the direction of tidal flow

he might have planned to leave for his trip home during low tide

Zielflug [23.3K]3 years ago
3 0
I would he would try to enter as the tide is rising, and leave as the tide is falling. Those things happen at all different times of day during a month. Hope it helps!!
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What is Cd2+ and F1- together as a formula
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What would diluting the vinegar do to the rate of a reaction between vinegar and sodium hydroxide? decrease the rate of reaction
GarryVolchara [31]
Vinegar is aceti acid diluted in water.

The reaction of vinegar with sodium hydroxide is:

CH3COOH + NaOH ---> CH3COONa + H2O

That is an acid-base reaction.

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Then, given thatn when you dilute the vinegar you decrease the concentration of CH3COOH, the rate of reaction will decrease.

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4 0
3 years ago
Hydrogen sulfide,H2S, is a very toxic gas with a smell of rotten eggs. using the following: H2S+3/2 O2=SO2+H2O H2+1/2O2=H2O S+O2
Serga [27]

Answer:

ΔH = -20kJ

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The enthalpy of formation of a compound is defined as the change of enthalpy during the formation of 1 mole of the substance from its constituent elements. For H₂S(g) the reaction that describes this process is:

H₂(g) + S(g) → H₂S(g)

Using Hess's law, it is possible to sum the enthalpies of several reactions to obtain the change in enthalpy of a particular reaction thus:

<em>(1) </em>H₂S(g) + ³/₂O₂(g) → SO₂(g) + H₂O(g) ΔH = -519 kJ

<em>(2) </em>H₂(g) + ¹/₂O₂(g) → H₂O(g) ΔH = -242 kJ

<em>(3) </em>S(g) + O₂(g) → SO₂(g) ΔH = -297 kJ

The sum of -(1) + (2) + (3) gives:

<em>-(1) </em>SO₂(g) + H₂O(g) → H₂S(g) + ³/₂O₂(g) ΔH = +519 kJ

<em>(2) </em>H₂(g) + ¹/₂O₂(g) → H₂O(g) ΔH = -242 kJ

<em>(3) </em>S(g) + O₂(g) → SO₂(g) ΔH = -297 kJ

<em>-(1) + (2) + (3): </em><em>H₂(g) + S(g) → H₂S(g) </em>

<em>ΔH =</em> +519kJ - 242kJ - 297kJ = <em>-20 kJ</em>

<em />

I hope it helps!

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4,1,5,3,2 (from left to right)

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