Answer:
Electromagnetic Waves
Explanation:
Electromagnetic waves do not need a medium to travel through, so therefore, they are faster.
They are located in Jerusalem.
Answer:
The answer is in the explanation
Explanation:
Acetic acid, CH₃COOH, is a weak acid that will produce a buffer when its conjugate base, CH₃COO⁻, acetate ion, is added to the solution.
<em>That is because a buffer is the mixture of a weak acid and its conjugate base or vice versa.</em>
When an acid (HX) is added to the solution, the acetate ion will react producing acetic acid, thus:
CH₃COO⁻ + HX → CH₃COOH + X⁻
For this reason, the pH doesn't change abruptly because H⁺ ions are not produced.
Now, if a base (BOH) is added to the buffer, CH₃COOH will react producing acetate ion and water, thus:
CH₃COOH + BOH → CH₃COO⁻ + H₂O + B⁺.
In the same way, there are not produced free OH⁻ and the pH doesn't change significantly.
Answer:
0.74 grams of methane
Explanation:
The balanced equation of the combustion reaction of methane with oxygen is:
it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.
firstly, we need to calculate the number of moles of both
for CH₄:
number of moles = mass / molar mass = (3.00 g) / (16.00 g/mol) = 0.1875 mol.
for O₂:
number of moles = mass / molar mass = (9.00 g) / (32.00 g/mol) = 0.2812 mol.
- it is clear that O₂ is the limiting reactant and methane will leftover.
using cross multiplication
1 mol of CH₄ needs → 2 mol of O₂
??? mol of CH₄ needs → 0.2812 mol of O₂
∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol
so 0.14 mol will react and the remaining CH₄
mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol
now we convert moles into grams
mass of CH₄ left over = no. of mol of CH₄ left over * molar mass
= 0.0469 mol * 16 g/mol = 0.7504 g
So, the right choice is 0.74 grams of methane
Answer:
A) 0.065 M is its molarity after a reaction time of 19.0 hour.
B) In 52 hours
will react 69% of its initial concentration.
Explanation:
![Co(NH_3)_5(H_2O)_3+[Co(NH_3)5Br]^{2+}(Purple)(aq)+H_2O(l)\rightarrow [Co(NH_3)_5(H_2O)]^{3+}(Pinkish-orange)(aq)+Br^-(aq)](https://tex.z-dn.net/?f=Co%28NH_3%29_5%28H_2O%29_3%2B%5BCo%28NH_3%295Br%5D%5E%7B2%2B%7D%28Purple%29%28aq%29%2BH_2O%28l%29%5Crightarrow%20%5BCo%28NH_3%29_5%28H_2O%29%5D%5E%7B3%2B%7D%28Pinkish-orange%29%28aq%29%2BBr%5E-%28aq%29)
The reaction is first order in
:
Initial concentration of
= ![[A_o]=0.100 M](https://tex.z-dn.net/?f=%5BA_o%5D%3D0.100%20M)
a) Final concentration of
after 19.0 hours= ![[A]](https://tex.z-dn.net/?f=%5BA%5D)
t = 19.0 hour = 19.0 × 3600 seconds ( 1 hour = 3600 seconds)
Rate constant of the reaction = k = 
The integrated law of first order kinetic is given as:
![[A]=[A_o]\times e^{-kt}](https://tex.z-dn.net/?f=%5BA%5D%3D%5BA_o%5D%5Ctimes%20e%5E%7B-kt%7D)
![[A]=0.100 M\times e^{-6.3\times 10^{-6} s^{-1}\times 19.0\times 3600 s}](https://tex.z-dn.net/?f=%5BA%5D%3D0.100%20M%5Ctimes%20e%5E%7B-6.3%5Ctimes%2010%5E%7B-6%7D%20s%5E%7B-1%7D%5Ctimes%2019.0%5Ctimes%203600%20s%7D)
![[A]=0.065 M](https://tex.z-dn.net/?f=%5BA%5D%3D0.065%20M)
0.065 M is its molarity after a reaction time of 19.0 h.
b)
Initial concentration of
= ![[A_o]=x](https://tex.z-dn.net/?f=%5BA_o%5D%3Dx)
Final concentration of
after t = ![[A]=(100\%-69\%) x=31\%x=0.31x](https://tex.z-dn.net/?f=%5BA%5D%3D%28100%5C%25-69%5C%25%29%20x%3D31%5C%25x%3D0.31x)
Rate constant of the reaction = k = 
The integrated law of first order kinetic is given as:
![[A]=[A_o]\times e^{-kt}](https://tex.z-dn.net/?f=%5BA%5D%3D%5BA_o%5D%5Ctimes%20e%5E%7B-kt%7D)

t = 185,902.06 s =
≈ 52 hours
In 52 hours
will react 69% of its initial concentration.