The answer is B. Suspension. Suspension mixtures are composed of two or more materials mixed together wherein the solute particles are usually larger than those found in a solution or colloid. In cases of solid-fluid suspension mixtures, the solid solute particles tend to settle at the bottom of the mixture after some time.
Answer:
2 moles of KCl will be produced
Explanation:
Given parameters:
Number of moles of K = 2 moles
Unknown:
Number of moles of KCl produced = ?
Solution:
To solve this problem;
Obtain a balanced chemical equation:
2K + Cl₂ → 2KCl ;
Since K is the limiting reactant, its amount will determine the extent of this reaction.
From the balanced equation;
2 moles of K will produce 2 moles of KCl
Given that 2 moles of K reacted, 2 moles of KCl will be produced
Answer:
<u>84.00 kPa = 630.084 torr</u>
Formula for kPa to torr: For an approximate result, multiply the pressure value by 7.501. <em><u>84* 7.501</u></em>
<u>84.00 kPa = 0.831683168 atm</u>
Formula for kPa to atm: for an approximate result, divide the pressure value by 101. <em><u>84/101</u></em>
When a electron moves from higher shell to lower shell ,it releases energy because if it goes to lower shell without releasing energy then it will unable to do that so.as lower electron shells have low energy electrons
<u>Answer:</u> The amount of Iodine-131 remain after 39 days is 0.278 grams
<u>Explanation:</u>
The equation used to calculate rate constant from given half life for first order kinetics:

where,
= half life of the reaction = 8.04 days
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 = 39 days
= initial amount of the sample = 8.0 grams
[A] = amount left after decay process = ?
Putting values in above equation, we get:
![0.0862=\frac{2.303}{39}\log\frac{8.0}{[A]}](https://tex.z-dn.net/?f=0.0862%3D%5Cfrac%7B2.303%7D%7B39%7D%5Clog%5Cfrac%7B8.0%7D%7B%5BA%5D%7D)
![[A]=0.278g](https://tex.z-dn.net/?f=%5BA%5D%3D0.278g)
Hence, the amount of Iodine-131 remain after 39 days is 0.278 grams