Im so sorry i don’t know I just need to answer 2 questions for me to be Able to ask
Considering that CCL3F gas behave like an ideal gas then we can use the Ideal Gas Law
<span>PV = nRT, however is an approximation and not the only way to resolve this problem with the given data..So,at the end of the solution I am posting some sources for further understanding and a expanded point of view. </span>
<span>Data: P= 856torr, T = 300K, V= 1.1L, R = 62.36 L Torr / KMol </span>
<span>Solving and substituting in the Gas equation for n = PV / RT = (856)(1.1L) /( 62.36)(300) = 0.05 Mol. This RESULT is of any gas. To tie it up to our gas we need to look for its molecular weight:MW of CCL3F = 137.7 gm/mol. </span>
<span>Then : 0.05x 137.5 = 6.88gm of vapor </span>
<span>If we sustract the vapor weight from the TOTAL weight of liquid we have: 11.5gm - 6.88gm = 4.62 gm of liquid.d</span>
Uranium emits particles and rays spontaneously through this process called radioactive decay or radioactivity.
Is the process by which the nucleus of an unstable atom loses energy by emitting radiation, including alpha particles, beta particles. gamma rays and conversion electrons.
Answer:
See explanation.
Explanation:
I highly suggest you watch OChem Tutor's videos on IUPAC nomenclature because the actual naming would take a lot of time to teach in text-based format. But here is how to name them:
1) I think there are two seperate pictures for number 1. The molecule on the left is 1-pentene and the one on the right is 4-methyl-1-pentene. If the whole thing is one molecule but there is just a bond missing where the red marker numbers are, that molecule would be 9-methyl-1,6-decadiene.
2) 4-methyl-2-pentene
3) 2,4-octadiene
4) 1,5-nonadiene
5) 2,5-dimethyl-3-hexene
6) 3,6-dimethyl-2,4-heptadiene
7) 2,5,5-trimethyl-2-hexene
3 minutes are left (16-13=3)