Ask question

Ask question

All categories

3 years ago

11

Please help!!!!! I need it!!!

1 answer:

jolli1 [7]3 years ago

7 0

It will be combustion of methane, because combustion will have the reactant and the oxygen, producing carbon dioxide and water, which is whats happening in this case.

You might be interested in

Baking soda and antacid tablets are examples of two common__________

attashe74 [19]

Bases

A base is a substance that dissociates into more hydroxide ions (-OH-) when dissolved in water. Bases are also good proton acceptors. Bases, therefore, reduce the number of H+ and increase OH- hence raising the pH of the solution.

B(aq) + H₂O(l) ⇌ BH⁺(aq) + OH⁻(aq)

Explanation:

Other properties of bases is that they are bitter to the taste and they feel slippery when touched. Strong bases are nonthlese very corrosive like acids. Bases turn red litmus paper blue. Most alkali hydroxides such as NaOH are bases.

Learn More:

For more on bases check out;

brainly.com/question/12574229

brainly.com/question/2015251

#LearnWithBrainly

7 0

3 years ago

Valerie is back in the lab to separate another blood sample, but her centrifuge has broken. Luckily, she remembers that red bloo

Alex

the answer is magnetic separation, not sedimentation separation

8 0

3 years ago

Read 2 more answers

Answer ASAP pls

Komok [63]

Answer:

wavelengths

Explanation:

4 0

2 years ago

The activation energy for a reaction is changed from 184 kJ/mol to 59.0 kJ/mol at 600. K by the introduction of a catalyst. If t

11111nata11111 [884]

Answer:

The catalyzed reaction will take 2.85 seconds to occur.

Explanation:

The activation energy of a reaction is given by:

$k = Ae^{-\frac{E_{a}}{RT}}$

For the reaction without catalyst we have:

$k_{1} = Ae^{-\frac{E_{a_{1}}}{RT}}$ (1)

And for the reaction with the catalyst:

$k_{2} = Ae^{-\frac{E_{a_{2}}}{RT}}$ (2)

Assuming that frequency factor (A) and the temperature (T) are constant, by dividing equation (1) with equation (2) we have:

$\frac{k_{1}}{k_{2}} = \frac{Ae^{-\frac{E_{a_{1}}}{RT}}}{Ae^{-\frac{E_{a_{2}}}{RT}}}$

$\frac{k_{1}}{k_{2}} = e^{\frac{E_{a_{2}} - E_{a_{1}}}{RT}$

$\frac{k_{1}}{k_{2}} = e^{\frac{59.0 \cdot 10^{3}J/mol - 184 \cdot 10^{3} J/mol}{8.314 J/Kmol*600 K} = 1.31 \cdot 10^{-11}$

Since the reaction rate is related to the time as follow:

$k = \frac{\Delta [R]}{t}$

And assuming that the initial concentrations ([R]) are the same, we have:

$\frac{k_{1}}{k_{2}} = \frac{\Delta [R]/t_{1}}{\Delta [R]/t_{2}}$

$\frac{k_{1}}{k_{2}} = \frac{t_{2}}{t_{1}}$

$t_{2} = t_{1}\frac{k_{1}}{k_{2}} = 6900 y*1.31 \cdot 10^{-11} = 9.04 \cdot 10^{-8} y*\frac{365 d}{1 y}*\frac{24 h}{1 d}*\frac{3600 s}{1 h} = 2.85 s$

Therefore, the catalyzed reaction will take 2.85 seconds to occur.

I hope it helps you!

4 0

3 years ago

Need a bf or gf fake or reallll<br> i dont care you can just take my points but .............. yea

Mrac [35]

Answer:

Hi

Explanation:

3 0

3 years ago

Read 2 more answers