Answer:
i want to help but i d k. the answer :( sorry i couldnt help
Explanation:
Start with the ideal gas equation, <span><span><span>PV=nRT</span> </span><span>PV=nRT</span></span>
and rearrange for pressure to get <span><span><span>p=<span><span>nRT</span>V </span></span> </span><span>p=<span><span>nRT</span>V</span></span></span>
. You have all the necessary variables in their proper units, so plug em' into the equation to solve for pressure in units of atmospheres.
<span><span><span>P=<span><span>(2.5 mol)(300 K)(0.08206 L atm <span><span>K </span><span><span>−1</span> </span></span> mo<span><span>l </span><span><span>−1</span> </span></span></span><span>50.0 L</span> </span>=1.23 atm</span> </span><span>P=<span><span>(2.5 mol)(300 K)(0.08206 L atm <span>K<span>−1</span></span> mo<span>l<span>−1</span></span></span><span>50.0 L</span></span>=1.23 atm</span></span>
All that needs to be done now is converting atmospheres to mm <span><span><span>Hg</span> </span><span>Hg</span></span>
.
<span><span><span>1.23 atm∗<span><span>760 mm Hg</span><span>1 atm</span> </span>=935 mm Hg</span> </span><span>1.23 atm∗<span><span>760 mm Hg</span><span>1 atm</span></span>=935 mm Hg</span></span>
.
That value makes sense, since the original pressure in atmospheres was above 1, the pressure in mm <span><span><span>Hg</span> </span><span>Hg</span></span>
will be above 760.
There are 11 electrons in Sodium(Na)
The first shell can hold 2 electrons
The second shell can hold 8 electrons
The third shell can also hold 8 electrons but there is only 1 left
so the answer is 2, 8, 1
Answer
She needs to set boundaries
Explanation:
Diet
Answer:
Potassium
Explanation:
In group one of the periodic table both the melting and the boiling points usually decrease down the group.
Now, down the group, it's lithium that comes first, then sodium, then potassium.
Thus, among the 3, potassium is furthermost down the group by virtue of the factor it has the highest atomic number.
Therefore, we can say that potassium has the lowest boiling point among the 3.
