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3 years ago

The abnormally high boiling point for water is due to...

1 answer:

3 0

Extensive hydrogen bonding.

It takes a lot of energy to break hydrogen bonds and turn liquid water into a gas, therefore the boiling point for water is high.

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The decomposition of hydrogen peroxide (H2O2(aq)) is proposed to follow a reaction mechanism of: Step 1: H2O2(aq) + Br–(aq)→H2O(

lys-0071 [83]

Answer:

Option (1) Br– is the catalyst, and the reaction follows a faster pathway with Br– than without

Explanation:

Let us consider the equation below:

Step 1:

H2O2(aq) + Br–(aq) → H2O(l) + BrO–(aq)

Step 2:

BrO–(aq) + H2O2(aq) → H2O(l) + O2(g) + Br–(aq)

From the above equation, we can see that Br– is unchanged.

This implies that Br– is the catalyst as catalyst does not take part in a chemical reaction but they create an alternate pathway to lower the activation energy in order for the reaction to proceed at a much faster rate to arrive at the products.

8 0

2 years ago

1 point If the boy is pushing with 50N of force and the static friction resistance is 70N of force, what will happen? *

solniwko [45]

Nothing, he shouldn’t be able to move it. Think about it like this say you try really hard to push something that is 5,000 pounds and you push as hard as you can. Well you can’t move it bc it weighs more than you can push. I’m sure their is a equation you can use to see how much you can push (body weight=force?)

6 0

3 years ago

Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$