Answer:
An inverse relationship can neither be represented by a straight line nor by a bar chart. But it can be represented by "xy = k"
Explanation:
Inverse relation is used for the values which are inversely related to each other. For example: Let suppose you have a value x and y. Then an increase in the value of x will result in the decrease of value y. Mathematically it is represented as,
x ∝ 1 / y
Where;
∝ = proportionality
Replacing the proportionality sign by a constant value "k" the relation becomes,
x = k / y
Solving for k,
x y = k
Conclusion:
Hence, an inverse relationship can be represented by "xy = k"
Answer:
No precipitate is formed.
Explanation:
Hello,
In this case, given the dissociation reaction of magnesium fluoride:
And the undergoing chemical reaction:
We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:
Next, the moles of magnesium chloride consumed by the sodium fluoride:
Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:
Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:
![[Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M](https://tex.z-dn.net/?f=%5BMg%5E%7B2%2B%7D%5D%3D%5Cfrac%7B3x10%5E%7B-4%7DmolMg%5E%7B%2B2%7D%7D%7B%280.3%2B0.5%29L%7D%20%3D3.75x10%5E%7B-4%7DM)
![[F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M](https://tex.z-dn.net/?f=%5BF%5E-%5D%3D%5Cfrac%7B2%2A3x10%5E%7B-4%7DmolMg%5E%7B%2B2%7D%7D%7B%280.3%2B0.5%29L%7D%20%3D7.5x10%5E%7B-4%7DM)
Thereby, the reaction quotient is:
In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.
Regards.
<h2>Question↷</h2>
Which of these solutions is the most basic?
[OH-] = 4.3 x 10-9 M
[OH-] = 2.8 x 10-11 M
[OH-] = 6.7 x 10-10 M
[OH-] = 1.0 x 10-5 M
<h2>Answer↷</h2>
- <u>[OH-] = 1.0 x 10-5 M</u> ✓
<h2>Solution↷</h2>
[OH-] = 4.3 x 10-9 M
- [OH-] = 4.3 x 10^-9 M
- pOH = - log [OH^1-]
- pOH = - log [4.3 x 10^-9]
- pOH = 8.36
- pH = 14- 8.36
- pH = 5.64
_______________________________________
[OH-] = 2.8 x 10-11 M
- [OH-] = 2.8 x 10^-11 M
- pOH = - log [OH^1-]
- pOH = - log [2.8 x 10^-11 ]
- pOH = 10.55
- pH = 14-10.55
- pH = 3.45
_______________________________________
[OH-] = 6.7 x 10-10 M
- [OH-] = 6.7 x 10^-10 M
- pOH = - log [OH^1-]
- pOH = - log [6.7 x 10^-10]
- pOH = 9.17
- pH = 14-9.17
- pH = 4.83
_______________________________________
[OH-] = 1.0 x 10-5 M
- [OH-] = 1.0 x 10-5 M
- pOH = - log [OH^1-]
- pOH = - log [1.0 x 10^-5 ]
- pOH = 5
- pH = 14-5
- pH = 9
_______________________________________
we know that , the solution with <u>pH > 7</u> is termed as basic and more the pH ,more the basicity, hence ,the solution with the highest pH would be the strongest base out of these all which is
<u>[OH-] = 1.0 x 10-5 M</u> with pH = 9
_______________________________________
<u>Answer:</u> The equation to calculate the mass of remaining isotope is ![[A]=\frac{20}{10^{-0.217t}}](https://tex.z-dn.net/?f=%5BA%5D%3D%5Cfrac%7B20%7D%7B10%5E%7B-0.217t%7D%7D)
<u>Explanation:</u>
The equation used to calculate rate constant from given half life for first order kinetics:
where,
= half life of the reaction = 
Putting values in above equation, we get:
Rate law expression for first order kinetics is given by the equation:
![k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B2.303%7D%7Bt%7D%5Clog%5Cfrac%7B%5BA_o%5D%7D%7B%5BA%5D%7D)
where,
k = rate constant = 
t = time taken for decay process
![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
= initial amount of the sample = 20 grams
[A] = amount left after decay process = ? grams
Putting values in above equation, we get:
![0.5=\frac{2.303}{t}\log\frac{20}{[A]}](https://tex.z-dn.net/?f=0.5%3D%5Cfrac%7B2.303%7D%7Bt%7D%5Clog%5Cfrac%7B20%7D%7B%5BA%5D%7D)
Hence, the equation to calculate the mass of remaining isotope is
