Answer:
The final volume in mL is 7.14 mL or 7.1 mL.
Explanation:
1.Use Boyle's Law(
). Re-arrange to solve for
<em> for the final volume.</em>
<em />
<em>2. Plug in values. </em>
The second option only.
<h3>Explanation</h3>
A base neutralizes an acid when the two reacts to produce water and a salt.
Sulfuric acid H₂SO₄ is the acid here. There are more than one classes of bases that can neutralize H₂SO₄. Among the options, there are:
Metal hydroxides
Metal hydroxides react with sulfuric acid to produce water and the sulfate salt of the metal.
.
The formula for calcium sulfate
in option A is spelled incorrectly. Why? The charge on each calcium
is +2. The charge on each sulfate ion
is -2. Unlike
ions, it takes only one
ion to balance the charge on each
ion. As a result,
and
ions in calcium sulfate exist on a 1:1 ratio.
.
Ammonia, NH₃
Ammonia NH₃ can also act as a base and neutralize acids. NH₃ exists as NH₄OH in water:
.
The ion
acts like a metal cation. Similarly to the metal hydroxides, NH₃ (or NH₄OH) neutralizes H₂SO₄ to produce water and a salt:
.
The formula of the salt (NH₄)₂SO₄ in the fourth option spelled the ammonium ion incorrectly.
As part of the salt (NH₄)₂SO₄, the ammonium ion NH₄⁺ is one of the products of this reaction and can't neutralize H₂SO₄ any further.
A heat because heat is the transfer of energy.
Answer:

Explanation:
The pressure at the bottom of the tank is:


The force exerted on the circular bottom is:
![F=(73581.921\,Pa)\cdot (\frac{\pi}{4} )\cdot [(12\,ft)\cdot (\frac{0.305\,m}{1\,ft} )]^{2}](https://tex.z-dn.net/?f=F%3D%2873581.921%5C%2CPa%29%5Ccdot%20%28%5Cfrac%7B%5Cpi%7D%7B4%7D%20%29%5Ccdot%20%5B%2812%5C%2Cft%29%5Ccdot%20%28%5Cfrac%7B0.305%5C%2Cm%7D%7B1%5C%2Cft%7D%20%29%5D%5E%7B2%7D)

Answer:

Explanation:
Hello there!
In this case, according to this question, we will need to deal with this dilution problem, because it is asking for the volume of a 12.1-M stock solution of HCl. In such a way, we can use the following equation, under the assumption of no change in the number of moles in the solution:

Thus, we solve for the initial volume, V1, as shown below:

And plug in the initial concentration and final concentration and volume to obtain:

Regards!