We use prefixes in ionic compounds Don't use numeric prefixes like mono, di, tri, etc. when naming ionic compounds - they are used only to denote covalent molecular compounds.
Answer:
0.1077 grams
Explanation:
First we will employ the ideal gas law to determine the number of moles of nitrogen gas.
PV=nRT
P=2 atm
V=20L
R=0.08206*L*atm*mol^-1*K^-1
T=323.15 K
Thus, 2atm*20L=n*0.08206*L*atm*mol^-1*K^-1*323.15K
K, atm, and L cancels out. Thus n=2*20mol/0.08206*323.15=1.5 moles
Lastly, we must convert the number of moles to grams. This can be done by dividing the number of moles by the molar mass of nitrogen gas, which is 14 grams.
1.5/14=0.1077 grams
Log base 10 of 1000000 is 6
Creativity allows a scientist to develop new ideas or evaluate old Ideas in a new way
:)
Answer:
a. P = 320 mmHg
Explanation:
For this problem, the pressure, volume, and temperature are changing, so we'll need to combine Boyle's Law and Charles' Law:
For this equation, the temperatures must be measured in Kelvin. The rest of units in the equation only need to match between beginning and end conditions.
Recall that to convert from Celsius to Kelvin, add 273, or use the equation 
.
So ![T_1=(44.4+273)[K]=317.4[K]](https://tex.z-dn.net/?f=T_1%3D%2844.4%2B273%29%5BK%5D%3D317.4%5BK%5D)
and ![T_2=(68.1+273)[K]=341.1[K]](https://tex.z-dn.net/?f=T_2%3D%2868.1%2B273%29%5BK%5D%3D341.1%5BK%5D)
Substituting known values, we can solve for the unknown:
![\dfrac{(574.6[mmHg])(208.1[mL])}{(317.4[K])}=\dfrac{P_2(401.5[mL])}{(341.1[K])}](https://tex.z-dn.net/?f=%5Cdfrac%7B%28574.6%5BmmHg%5D%29%28208.1%5BmL%5D%29%7D%7B%28317.4%5BK%5D%29%7D%3D%5Cdfrac%7BP_2%28401.5%5BmL%5D%29%7D%7B%28341.1%5BK%5D%29%7D)
![\dfrac{(574.6[mmHg])(208.1[mL]\!\!\!\!\!\!\!\!\!\!{--} )}{317.4[K]\!\!\!\!\!\!{-}}*\dfrac{341.1[K] \!\!\!\!\!\!{-}}{401.5[mL] \!\!\!\!\!\!\!\!\!\!\!{--} }=\dfrac{P_2(401.5[mL] \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{-----})}{341.1[K] \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----}}*\dfrac{341.1[K] \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----}}{401.5[mL] \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{-----} }](https://tex.z-dn.net/?f=%5Cdfrac%7B%28574.6%5BmmHg%5D%29%28208.1%5BmL%5D%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%7B--%7D%20%29%7D%7B317.4%5BK%5D%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%7B-%7D%7D%2A%5Cdfrac%7B341.1%5BK%5D%20%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%7B-%7D%7D%7B401.5%5BmL%5D%20%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%7B--%7D%20%7D%3D%5Cdfrac%7BP_2%28401.5%5BmL%5D%20%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%7B-----%7D%29%7D%7B341.1%5BK%5D%20%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%7B----%7D%7D%2A%5Cdfrac%7B341.1%5BK%5D%20%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%7B----%7D%7D%7B401.5%5BmL%5D%20%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%7B-----%7D%20%7D)
![320.056719296[mmHg]=P_2](https://tex.z-dn.net/?f=320.056719296%5BmmHg%5D%3DP_2)
Accounting for significant figures, ![P_2=320.1[mmHg]](https://tex.z-dn.net/?f=P_2%3D320.1%5BmmHg%5D)
.
The closest answer provided is 320, so "a".
