Answer:
The intermolecular forces between water molecules are stronger than those between oxygen molecules. In general, the bigger the molecule, the stronger the intermolecular forces, so the higher the melting and boiling points.
Answer: The correct answer is option B.
Explanation: There are 7 elements in group VIII-A. They are Helium (He), Neon(Ne), Argon(Ar), Krypton (Kr), Xenon (Xe), Radon (Rn) and Ununoctium (Uuo). They are considered as a part of group 18.
Metals are shown in purple color in the periodic table. Metalloids as blue color and Non-metals are shown in green color.
Group VIII-A elements have fully filled orbitals and hence are most stable amongst all the elements in the periodic table. They are considered as least reactive elements and as they are categorized as non-metals. So, they are the least reactive non-metals.
Therefore, the correct option is B.
Answer:
50/36 = 25/18
Explanation:
Solution at attachment box
Molality = mole of dissolvable (this question glucose) / kg of water
Answer:
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Answer:
final volume = 10.5 Liters N₂(g) at 21°C and 823Torr*
Explanation:
*Note=>No specified mass value of N₂(g) is defined in the problem. Therefore for a starting point, the gas sample is assumed to be 1.00 mole N₂(g) at STP conditions 22.4L
Determine volume of N₂(g) at 21°C(=294K) and 823 Torr (= 2.286 Atm).
Start with Volume of N₂(g) at 0°C and 1 Atm pressure => 22.4L and adjust to final volume of N₂(g) based upon 21°C(=294K) and 823 Torr (= 2.286 Atm).
V(final) = 22.4L(294K/273K)(360 Torr/823 Torr) = 22.4L(294/273)(360/823) = 10.55 Liters final volume.
Note: The volume of 1 mole (assumed) of any gas at STP (0°C/1 Atm) is 22.4 Liters. To convert to non-STP conditions, convert temperature and pressure factors (changes) that reflect what happens when the gas is expanded or decreased; but, these adjustments are taken independently for each variable of interest. The following notes explain.
For the increase in temperature from 0°C(=273K) to 21°C(=294K) one must apply a temperature ratio that will increase volume. That is, the change in volume due to the temperature change is 294K/273K. If a 273K/294K ratio were used the volume would have decreased. Not so for heating a sample of gas.
For the increase in pressure one should expect a decrease in volume. Therefore apply a pressure ration that will effectively decrease the volume of the gas. That is, to decrease a 22.4L sample at STP multiply the standard volume by a ratio of pressures that will decrease 22.4L to a smaller volume. That is, V(final by pressure effects) multiply by 360Torr/823Torr to decrease the STP VOLUME (22.4L) to the new non-standard volume. If 823Torr/360Torr were used, the final volume would not be smaller, but larger. Such is the physical effect of an increasing pressure change.