Answer:
Option A.
2Na + 2H2O —> 2NaOH + H2
Explanation:
To know which option is correct, we shall do a head count of the number of atoms present on both side to see which of them is balanced. This is illustrated below below:
For Option A:
2Na + 2H2O —> 2NaOH + H2
Reactant >>>>>>> Product
2 Na >>>>>>>>>>> 2 Na
4 H >>>>>>>>>>>> 4 H
2 O >>>>>>>>>>>> 2 O
Thus, the above equation is balanced.
For Option B:
2Na + 2H2O —> NaOH + H2
Reactant >>>>>>> Product
2 Na >>>>>>>>>>> 1 Na
4 H >>>>>>>>>>>> 3 H
2 O >>>>>>>>>>>> 1 O
Thus, the above equation is not balanced.
For Option C:
2Na + H2O —> 2NaOH + H2
Reactant >>>>>>> Product
2 Na >>>>>>>>>>> 2 Na
2 H >>>>>>>>>>>> 4 H
1 O >>>>>>>>>>>> 2 O
Thus, the above equation is not balanced.
For Option D:
Na + 2H2O —> NaOH + 2H2
Reactant >>>>>>> Product
1 Na >>>>>>>>>>> 1 Na
4 H >>>>>>>>>>>> 5 H
2 O >>>>>>>>>>>> 1 O
Thus, the above equation is not balanced.
From the illustrations made above, only option A is balanced.
Answer:
Melting butter
Explanation:
You can reverse the change of butter back to its original state but you can never reverse the rest back to there original state
CH₇ is the empirical formula of the car fuel.
Explanation:
To find the empirical formula we use the following algorithm.
First divide each mass the the molar weight of each element:
for carbon 2.87 / 12 = 0.239
for hydrogen 3.41 / 2 = 1.705
And now divide each quantity by the lowest number which is 0.239:
for carbon 0.239 / 0.239 = 1
for hydrogen 1.705 / 0.239 = 7.13 ≈ 7
The empirical formula of the car fuel is CH₇.
I have to tell you that in reality this formula is wrong because is not possible to exist. However the algorithm for finding the empirical formula is right, the problem may reside in the amounts of carbon and hydrogen given.
Learn more about:
empirical formula
brainly.com/question/5297213
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567000 would be in grams, hope this helps :)
Answer:
(a) 7.11x10⁻⁴ M/s
(b) 2.56 mol.L⁻¹.h⁻¹
Explanation:
(a) The reaction is:
O₃(g) + NO(g) → O₂(g) + NO₂(g) (1)
The reaction rate of equation (1) is given by:
![rate = k*[O_{3}][NO]](https://tex.z-dn.net/?f=%20rate%20%3D%20k%2A%5BO_%7B3%7D%5D%5BNO%5D%20)
(2)
<u>We have:</u>
k: is the rate constant of reaction = 3.91x10⁶ M⁻¹.s⁻¹
[O₃]₀ = 2.35x10⁻⁶ M
[NO]₀ = 7.74x10⁻⁵ M
Hence, to find the inital reacion rate we will use equation (2):
![rate = k*[O_{3}]_{0}[NO]_{0} = 3.91 \cdot 10^{6} M^{-1}s^{-1}*2.35\cdot 10^{-6} M*7.74 \cdot 10^{-5} M = 7.11 \cdot 10^{-4} M/s](https://tex.z-dn.net/?f=%20rate%20%3D%20k%2A%5BO_%7B3%7D%5D_%7B0%7D%5BNO%5D_%7B0%7D%20%3D%203.91%20%5Ccdot%2010%5E%7B6%7D%20M%5E%7B-1%7Ds%5E%7B-1%7D%2A2.35%5Ccdot%2010%5E%7B-6%7D%20M%2A7.74%20%5Ccdot%2010%5E%7B-5%7D%20M%20%3D%207.11%20%5Ccdot%2010%5E%7B-4%7D%20M%2Fs%20)
Therefore, the inital reaction rate is 7.11x10⁻⁴ M/s
(b) The number of moles of NO₂(g) produced per hour per liter of air is:
t = 1 h
V = 1 L
![\frac{\Delta[NO_{2}]}{\Delta t} = rate](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%5BNO_%7B2%7D%5D%7D%7B%5CDelta%20t%7D%20%3D%20rate)
![\frac{\Delta[NO_{2}]}{\Delta t} = 7.11 \cdot 10^{-4} M/s*\frac{3600 s}{1 h} = 2.56 mol.L^{-1}.h{-1}](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%5BNO_%7B2%7D%5D%7D%7B%5CDelta%20t%7D%20%3D%207.11%20%5Ccdot%2010%5E%7B-4%7D%20M%2Fs%2A%5Cfrac%7B3600%20s%7D%7B1%20h%7D%20%3D%202.56%20mol.L%5E%7B-1%7D.h%7B-1%7D)
Hence, the number of moles of NO₂(g) produced per hour per liter of air is 2.56 mol.L⁻¹.h⁻¹
I hope it helps you!
