<h3>
Answer:</h3>
The centripetal acceleration is 26.38 m/s²
<h3>
Explanation:</h3>
We are given;
- Mass of rubber stopper = 13 g
- Length of the string(radius) = 0.93 m
- Time for one revolution = 1.18 seconds
We are required to calculate the centripetal acceleration.
To get the centripetal acceleration is given by the formula;
Centripetal acc = V²/r
Where, V is the velocity and r is the radius.
Since time for 1 revolution is 1.18 seconds,
Then, V = 2πr/t, taking π to be 3.142 ( 1 revolution = 2πr)
Therefore;
Velocity = (2 × 3.142 × 0.93 m) ÷ 1.18 sec
= 4.953 m/s
Thus;
Centripetal acceleration = (4.953 m/s)² ÷ 0.93 m
= 26.38 m/s²
Hence, the centripetal acceleration is 26.38 m/s²
Answer:
Equilibrium shifts to the right
Explanation:
An exothermic reaction is one in which temperature is released to the environment. Hence, if the reaction vessel housing an exothermic reaction is touched after reaction completion, we will notice that the reaction vessel e.g beaker is hot.
To consider the equilibrium response to temperature changes, we need to consider if the reaction is exothermic or endothermic. In the case of this particular question, it has been established that the reaction is exothermic.
Heat is released to the surroundings as the reactants are at a higher energy level compared to the products. Hence, increasing the temperature will favor the formation of more reactants and as such, the equilibrium position will shift to the left to pave way for the formation of more reactants. Thus , more acetylene and hydrogen would be yielded
Answer:
I think <em><u>alpha</u></em> and <em><u>beta</u></em> is the answer.
Answer:
11.39
Explanation:
Given that:
Given that:
Mass = 1.805 g
Molar mass = 82.0343 g/mol
The formula for the calculation of moles is shown below:
Thus,
Given Volume = 55 mL = 0.055 L ( 1 mL = 0.001 L)
Concentration = 0.4 M
Consider the ICE take for the dissociation of the base as:
B + H₂O ⇄ BH⁺ + OH⁻
At t=0 0.4 - -
At t =equilibrium (0.4-x) x x
The expression for dissociation constant is:
![K_{b}=\frac {\left [ BH^{+} \right ]\left [ {OH}^- \right ]}{[B]}](https://tex.z-dn.net/?f=K_%7Bb%7D%3D%5Cfrac%20%7B%5Cleft%20%5B%20BH%5E%7B%2B%7D%20%5Cright%20%5D%5Cleft%20%5B%20%7BOH%7D%5E-%20%5Cright%20%5D%7D%7B%5BB%5D%7D)
x is very small, so (0.4 - x) ≅ 0.4
Solving for x, we get:
x = 2.4606×10⁻³ M
pOH = -log[OH⁻] = -log(2.4606×10⁻³) = 2.61
<u>pH = 14 - pOH = 14 - 2.61 = 11.39</u>
