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noname [10]
3 years ago
8

HURRY !!!!!! How did kelper describe the planets Orbits ?

Chemistry
1 answer:
Andrei [34K]3 years ago
8 0
Kepler's Laws of Planetary Motion. While Copernicus rightly observed that the planets revolve around the Sun, it was Kepler who correctly defined their orbits.

Sorry about the holds, copied it from google...
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Nina [5.8K]
I think 3.33 maybe ‍♀️
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PLEASE HELP i’ll mark brainliest ( reporting fake answers)
kherson [118]

Delta S reaction= Delta S products- Delta S reactants

don't forget to mulitiply by coefficients

also

here is a really slow way to do it

you know the moles of gas increased

so Delta S is positive

so its B or D

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6 0
3 years ago
The reactivity of metals on the periodic table
kari74 [83]

Explanation:

The most reactive metals are found on the left of the periodic table, in the blue column, known as the alkali metals. Their reactivity increases as we go down column (group) one. Reactive metals, when attached to less reactive metals, have the ability to prevent the less reactive metal from rusting.

8 0
3 years ago
2/10
emmainna [20.7K]

Answer:

pH=2.34

Explanation:

HBr  -> H  + Br

The dissociation it's complete, for that reason the concentration of the products is the same of HBr

[H+]=[Br-]=0.00234 M

pH= - log (0.00234)=2.34

6 0
3 years ago
Consider the following mechanism for the oxidation of bromide ions by hydrogen peroxide in aqueous acid solution. H+ + H2O2 ? H3
Margarita [4]

<u>Answer:</u> The rate law for the reaction is \text{Rate}=k'[H+][H_2O_2][Br^-]

<u>Explanation:</u>

Rate law is the expression which is used to express the rate of the reaction in terms of the molar concentration of reactants where each term is raised to the power their stoichiometric coefficient respectively from a balanced chemical equation.

In a mechanism of the reaction, the slow step in the mechanism determines the rate of the reaction.

The chemical equation for the oxidation of bromide ions by hydrogen peroxide in aqueous acid solution follows:

2H^++2Br^-+H_2O_2\rightarrow Br_2+2H_2O

The intermediate reaction of the mechanism follows:

<u>Step 1:</u>  H^++H_2O_2\rightleftharpoons H_3O_2^+;\text{ (fast)}

<u>Step 2:</u>  H_3O_2^++Br^-\rightarrow HOBr+H_2O;\text{(slow)}

<u>Step 3:</u>  HOBr+H^++Br^-\rightarrow Br_2+H_2O;\text{(fast)}

As, step 2 is the slow step. It is the rate determining step

Rate law for the reaction follows:

\text{Rate}=k[H_3O_2^+][Br^-]          ......(1)

As, [H_3O_2^+] is not appearing as a reactant in the overall reaction. So, we apply steady state approximation in it.

Applying steady state approximation for [H_3O_2^+] from step 1, we get:

K=\frac{[H_3O_2^+]}{[H^+][H_2O_2]}  

[H_3O_2^+]=K[H^+][H_2O_2]

Putting the value of [H_3O_2^+] in equation 1, we get:

\text{Rate}=k.K[H^+][H_2O_2][Br^-]\\\\\text{Rate}=k'[H+][H_2O_2][Br^-]

Hence, the rate law for the reaction is \text{Rate}=k'[H+][H_2O_2][Br^-]

4 0
3 years ago
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