Answer:
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Explanation:
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The mass of a molecule is very very small but there is many molecules and they all have kinetic energy. First, we have to assume that the gas is ideal. Kinetic energy of an ideal gas is only a function of moles and temperature. Hence:
KE=(3/2)nRT = (3/4)(1mole)(8.314J/molK)(25+273.15)K
= 3718.22J or 3.7KJ
ANSWER IS A true
because both ch3oh and water has hydrogen bonding
<u>Solution and Explanation:</u>
- Strong peak at 643cm−1 corresponds to alkene C=C stretching.
Medium peak at 3077cm−1 corresponds =C-H stretching.
Strong peaks at 912cm−1 and 994cm−1 corresponds to alkene =C-H bending. This tells the alkene is monosubstituted.
2. Weak peak at 3077cm−1 corresponds =C-H stretching.
The compound shows strong peak at 833cm−1 corresponds to alkene =C-H bending and weak peak at 1667cm−1 corresponds to alkene C=C stretching. This tells the alkene is trisubstituted.
3. Weak peak at 1665cm−1 corresponds to alkene C=C stretching.
Medium peak at 3086cm−1 corresponds =C-H stretching.
Strong peaks at 714cm−1 corresponds to alkene =C-H bending. This tells the alkene is cis-disubstituted.
4. Medium peak at 1643cm−1 corresponds to alkene C=C stretching.
Medium peak at 3010cm−1 corresponds =C-H stretching.
Strong peaks at 885cm−1 corresponds to alkene =C-H bending. This tells the alkene is gem-dialkylsubstituted.
5. Medium peak at 1650cm−1 corresponds to alkene C=C stretching.
Medium peak at 3040cm−1 corresponds =C-H stretching.
Strong peaks at 967cm−1 corresponds to alkene =C-H bending. This tells the alkene is trans-disubstituted.
The answer is 60.3% magnesium, 39.7% oxygen.
Solution:
The chemical equation for the reaction is 2 Mg + O2 → 2 MgO.
Since magnesium reacts completely with oxygen, it is the limiting reactant in the reaction. Hence, we can use the number of moles of magnesium to get the mass of MgO produced:
moles of magnesium = 14.7g / 24.305g mol-1
= 0.6048 mol
mass of MgO = 0.6048mol Mg(2 mol MgO/2mol Mg)(40.3044g MgO/1 mol MgO)
= 24.376g MgO
We can now solve for the percentage of magnesium:
% Mg = (14.7g Mg / 24.376g MgO)*100% = 60.3%
We also use the number of moles of magnesium to get the mass of oxygen consumed in the reaction:
mass of O2 = 0.6048 mol Mg (1mol O2 / 2mol Mg) (31.998g / 1mol O2)
= 9.676g
The percentage of oxygen is therefore
% O2 = (9.676g O2 / 24.376g MgO)*100%
= 39.7%
Notice that we can just subtract the magnesium's percentage from 100% to get
% O2 = 100% - 60.3% = 39.7%