Answer:
6.2 calories
Explanation:
Data Given:
change in temperature = 20 °C
specific heat of gold = 0.031 calories/gram °C
mass of gold = 10.0 grams
Amount of Heat = ?
Solution:
Formula used
Q = Cs.m.ΔT
Where:
Q = amount of heat
Cs = specific heat of gold = 0.031 calories/gram °C
m = mass
ΔT = Change in temperature
Put values in above equation
Q = 0.031 calories/gram °C x 10.0 g x 20 °C
Q = 6.2 calories
So option A is correct = 6.2 calories
Answer:
2 Answers. The column is filled with the carrier (liquid or gas) before the sample is injected. Thus if there is no interaction between the sample and the column, then the fastest that the sample can get to the detector is the dead time denoted by tM in the diagram.
Answer:
Noble gas, any of the seven chemical elements that make up Group 18 (VIIIa) of the periodic table. The elements are helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), radon (Rn), and oganesson (Og).
Explanation:
Noble gas, any of the seven chemical elements that make up Group 18 (VIIIa) of the periodic table. The elements are helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), radon (Rn), and oganesson (Og).
Answer: Heat of vaporization is 41094 Joules
Explanation:
The vapor pressure is determined by Clausius Clapeyron equation:
where,
= initial pressure at 429 K = 760 torr
= final pressure at 415 K = 515 torr
= enthalpy of vaporisation = ?
R = gas constant = 8.314 J/mole.K
= initial temperature = 429 K
= final temperature = 515 K
Now put all the given values in this formula, we get
![\log (\frac{515}{760}=\frac{\Delta H}{2.303\times 8.314J/mole.K}[\frac{1}{429K}-\frac{1}{415K}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7B515%7D%7B760%7D%3D%5Cfrac%7B%5CDelta%20H%7D%7B2.303%5Ctimes%208.314J%2Fmole.K%7D%5B%5Cfrac%7B1%7D%7B429K%7D-%5Cfrac%7B1%7D%7B415K%7D%5D)
Thus the heat of vaporization is 41094 Joules
Full question options;
(Fe, Pb, Mg, or Ca)
Answer:
Iron - Fe
Explanation:
We understand tht metals pretty much form bonds by losing their valence (outermost electrons). But this question specifically asks for metals that lose beyond their outermost electrons; next to outermost principal energy levels.
Pb, Mg, and Ca only lose their outermost electrons to form the following ions;
Pb2+, Mg2+, and Ca2+.
This is because their ions have achieved a stable octet configuration - the dreamland of atoms where they are satisfied and don't need to go into reactions again.
Iron on the other hand has the following electronic configurations;
Fe: [Ar]4s2 3d6
Fe2+: [Ar]4s0 3d6
Fe3+: [Ar]4s0 3d5
This means ion can lose both the ooutermost electrons (4s) and next to outermost principal energy levels (3d). So correct option is Iron.
