Answer:
It will take 188.06 hours for the concentration of A to decrease 10.0% of its original concentration.
Explanation:
A → B
Initial concentration of the reactant = x
Final concentration of reactant = 10% of x = 0.1 x
Time taken by the sample, t = ?
Formula used :
where,
= initial concentration of reactant
A = concentration of reactant left after the time, (t)
= half life of the first order conversion = 56.6 hour
= rate constant
Now put all the given values in this formula, we get
t = 188.06 hour
It will take 188.06 hours for the concentration of A to decrease 10.0% of its original concentration.
<u>Answer:</u>
The correct answer option is B) 2.0 M.
<u>Explanation:</u>
We are given the number of grams of NaOH (Sodium Chloride) which are dissolved in 750 milliliters of water and we are to find its molarity.
We know the formula of molarity:
<em>Molarity = (mass * 1000) / (volume * molecular mass) </em>
Volume = 750 ml = 750 cm
Molecular mass = 40
Mass = 60 grams
Substituting these values in the above formula:
Molarity = 
= 2.0 M
Answer
Using the law of reflection—the angle of reflection equals the angle of incidence—we can see that the image and object are the same distance from the mirror. This is a virtual image, since it cannot be projected—the rays only appear to originate from a common point behind the mirror.
Explanation:
Hope this helps someone
Answer:
C8H8 + 10O2 → 8CO2 + 4H2O
Explanation:
unbalanced reaction:
C8H8 + O2 → CO2 + H2O
balanced for semireactions:
(1) 16H2O + C8H8 → 8CO2 + 40H+
(2) 10(4H+ + O2 → 2H2O)
⇒ 40H+ + 10O2 → 20H2O
(1) + (2):
balanced reaction:
⇒ C8H8 + 10O2 → 8CO2 + 4H2O
8 - C - 8
20 - O2 - 20
8 - H - 8
<h3>
Answer:</h3>
8 alpha particles
4 beta particles
<h3>
Explanation:</h3>
<u>We are given;</u>
- Neptunium-237
- Thallium-205
- Neptunium-237 undergoes beta and alpha decay to form Thallium-205.
We are required to determine the number of beta and alpha particles produced to complete the decay series.
- We need to know that when a radioisotope emits an alpha particle the mass number reduces by 4 while the atomic number decreases by 2.
- When a beta particle is emitted the mass number of the radioisotope increases by 1 while the atomic number remains the same.
In this case;
Neptunium-237 has an atomic number 93, while,
Thallium-205 has an atomic number 81.
Therefore;
²³⁷₉₃Np → x⁴₂He + y⁰₋₁e + ²⁰⁵₈₁Ti
We can get x and y
237 = 4x + y(0) + 205
237-205 = 4x
4x = 32
x = 8
On the other hand;
93 = 2x + (-y) + 81
but x = 8
93 = 16 -y + 81
y = 4
Therefore, the complete decay equation is;
²³⁷₉₃Np → 8⁴₂He + 4⁰₋₁e + ²⁰⁵₈₁Ti
Thus, Neptunium-237 emits 8 alpha particles and 4 beta particles to become Thallium-205.
