Answer:
You need to add 400mL of water
Explanation:
500mL = 5 M HCI That means that if you divide both sides by 5
100mL = 1 M HCI If you need ot get rid of 4 M HCI then you add 400 mL of water because that is what it is equal to
Answer:
The answer is
<h2>59.6 g </h2>
Explanation:
The mass of a substance when given the density and volume can be found by using the formula
<h3>mass = Density × volume</h3>
From the question
Density of aluminum = 2.00 g/mL
volume = 29.8 mL
The mass is
mass = 2 × 29.8
We have the final answer as
<h3>59.6 g</h3>
Hope this helps you
Correct Answer: Option C
Reason:
<span>The </span>Pauli Exclusion Principle<span> states as '<em>in an atom or molecule, no two electrons can have the same four electronic quantum numbers. Further, an orbital can contain a maximum of only two electrons, the two electrons must have opposing spins.</em>'
</span>
Thus, it can be seen that in option C, electrons in last 2 subshell have electrons with same spin, which is a violation of Pauli Exclusion Principle .
Answer:
Choice d. No effect will be observed as long as other factors (temperature, in particular) are unchanged.
Explanation:
The equilibrium constant of a reaction does not depend on the pressure. For this particular reaction, the equilibrium quotient is:
![Q = \displaystyle \frac{[\mathrm{SO_2\, (g)}]}{[\mathrm{O_2\, (g)}]}](https://tex.z-dn.net/?f=Q%20%3D%20%5Cdisplaystyle%20%5Cfrac%7B%5B%5Cmathrm%7BSO_2%5C%2C%20%28g%29%7D%5D%7D%7B%5B%5Cmathrm%7BO_2%5C%2C%20%28g%29%7D%5D%7D)
.
Note that the two sides of this balanced equation contain an equal number of gaseous particles. Indeed, both ![[\mathrm{SO_2\, (g)}]](https://tex.z-dn.net/?f=%5B%5Cmathrm%7BSO_2%5C%2C%20%28g%29%7D%5D)
and ![[\mathrm{O_2\, (g)}]](https://tex.z-dn.net/?f=%5B%5Cmathrm%7BO_2%5C%2C%20%28g%29%7D%5D)
will increase if the pressure is increased through compression. However, because 
and 
have the same coefficients in the equation, their concentrations are raised to the same power in the equilibrium quotient 
.
As a result, the increase in pressure will have no impact on the value of 
. If the system was already at equilibrium, it will continue to be at an equilibrium even after the change to its pressure. Therefore, no overall effect on the equilibrium position should be visible.
