Explanation:
Take shelter in a hard wall building
Close doors and windows cut off ventilation
Not 100% sure if its right.
When an object covers equal distances in equal amounts of time, it is moving at a constant speed.
There will be needed 982.35 mL of solution to obtain 16.1 grams of the salt.There will be needed mL of
Why?
In order to calculate how many milliliters are needed to obtain 16.1 grams of the salt given its concentration, we first need to find its chemical formula which is the following:

Now that we know the chemical formula of the substance, we need to find its molecular mass. We can do it by the following way:

We have that the molecular mass of the substance will be:

Therefore, knowing the molecular mass of the substance, we need to calculate how many mols represents 16.1 grams of the same substance, we can do it by the following way:


Finally, if we need to calculate how many milliliters are needed, we need to use the following formula:


Now, substituting and calculating, we have:

Henc, there will be needed 982.35 mL of solution to obtain 16.1 grams of the salt.
Have a nice day!
Answer:
0.22 mol HClO, 0.11mol HBr.
0.25mol NH₄Cl, 0.12 mol HCl
Explanation:
A buffer is defined as a mixture in solution between weak acid and its conjugate base or vice versa.
Potassium hypochlorite (KClO) could be seen as conjugate base of HClO (Weak acid). That means the addition of <em>0.22 mol HClO </em>will convert the solution in a buffer. HBr reacts with KClO producing HClO, thus, <em>0.11mol HBr</em> will, also, convert the solution in a buffer. 0.23 mol HBr will react completely with KClO and in the solution you will have only HClO, no a buffering system.
Ammonia (NH₃) is a weak base and its conjugate base is NH₄⁺. That means the addition of <em>0.25mol NH₄Cl</em> will convert the solution in a buffer. Also, NH₃ reacts with HCl producing NH₄⁺. Thus, addition of<em> 0.12 mol HCl</em> will produce NH₄⁺. 0.25mol HCl consume all NH₃.
Answer:
vineger
Explanation:
homogeneous mixture cannot be separated pyhsically