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arlik [135]
2 years ago
13

What would the rate law be for the following reaction:

Chemistry
1 answer:
lana [24]2 years ago
6 0

Answer:

C. Rate = k[H2]^2[O2]

Explanation:

Rate law only cares about REACTANTS. Since, rate law can only be determined experimentally, I am assuming the given reaction mechanism is elementary reaction from which we can write the rate law.

Only H2 and O2 are part of rate law since they are reactants and also the coefficient in front of H2 goes as exponent on rate law to indicate the order of H2 in the reaction.

Rate= k [H2]^2 [O2]

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What occurs in a chemical reaction?
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B - The reactants are the starting substances and the products are the end substances.
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2 years ago
Read 2 more answers
Bromine can be classified as a
sergiy2304 [10]

Answer:

Halogen / salt-former

Explanation:

Bromine is classified as an element in the 'Halogens' section which can be located in group 7 of the Periodic Table. The term "halogen" means "salt-former" and compounds containing halogens are called "salts".

8 0
3 years ago
Why do you need many different indicators to span the entire ph spectrum?
marin [14]
The answer is each indicator has a narrow range.  We need many different indicators to span the entire ph spectrum because each indicator has a narrow range.
4 0
3 years ago
If a piece of cadmium with a mass of 37.60 g and a temperature of 100.0 oC is dropped into 25.00 cc of water at 23.0 oC, what wi
zlopas [31]

Answer:

T_{eq}=28.9\°C

Explanation:

Hello!

In this case, since it is observed that hot cadmium is placed in cold water, we can infer that the heat released due to the cooling of cadmium is gained by the water and therefore we can write:

Q_{Cd}+Q_{W}=0

Thus, we insert mass, specific heat and temperatures to obtain:

m_{Cd}C_{Cd}(T_{eq}-T_{Cd})+m_{W}C_{W}(T_{eq}-T_{W})=0

In such a way, since the specific heat of cadmium and water are respectively 0.232 and 4.184 J/(g °C), we can solve for the equilibrium temperature (the final one) as shown below:

T_{eq}=\frac{m_{Cd}C_{Cd}T_{Cd}+m_{W}C_{W}T_{W}}{m_{Cd}C_{Cd}+m_{W}C_{W}}

Now, we plug in to obtain:

T_{eq}=\frac{37.60g*0.232\frac{J}{g\°C}*100.00\°C+25.00g*4.184\frac{J}{g\°C}*23.0\°C}{37.60g*0.232\frac{J}{g\°C}+25.00g*4.184\frac{J}{g\°C}}\\\\T_{eq}=28.9\°C

NOTE: since the density of water is 1g/cc, we infer that 25.00 cc equals 25.00 g.

Best regards!

6 0
2 years ago
Automobile bodies contain significant amounts of iron. The iron is protected by the addition of zinc. This is called galvanizati
Oksana_A [137]

<u>Answer:</u> The balanced chemical equation is written below.

<u>Explanation:</u>

Galvanization is defined as the process in which a protective layer of zinc is applied to iron or steel to prevent the metal from rusting.

Zinc prevents the oxidation of iron and acts as a reducing agent in the process.

The half reaction for the process follows:

<u>Oxidation half reaction:</u>  Zn\rightarrow Zn^{2+}+2e^-

<u>Reduction half reaction:</u>  Fe^{2+}+2e^-\rightarrow Fe

Net chemical equation:  Zn+Fe^{2+}\rightarrow Zn^{2+}+Fe

Hence, the balanced chemical equation is written above.

6 0
3 years ago
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