Answer:
methanol because it is liquid at room temperature which means that it melts at a lower temperature
Answer:
I, III, and V
Explanation:
All five options are correct.
However, your instructor is probably expecting the answer I, III, and V.
These three elements are on the left-hand side of the Periodic Table. Thus, they tend to lose electrons and form positive ions most easily.
H is the most reluctant to form a cation. Its first ionization energy is 1100 kJ·mol⁻¹.
However, F and He can lose electrons to form cations.
It takes only 1700 kJ·mol⁻¹ to remove an electron from F. However, F tends to add electrons to get a complete valence shell.
Even He, with a complete valence shell, will give up a valence electron for the expenditure of only 2400 kJ·mol⁻¹.
Correct Answer: CH2CCI2 and CH2CH2.
Reason:
Structure of compound depends on the hybridization of central atom. In both of the above molecules (i.e. CH2CCL2 and CH2CH2). Central atom i.e. carbon is
sp^2 hybridized. Thus 3 hybridized orbitals of 'C' is involved in sigma bonding. Further, these orbitals are placed spatially at an angle of
. The un-hybridized 2pz orbital of 'C' is involved in pi-bonding. Due to this, both of the above molecules possess a
triangular planner geometry.
Answer:
0.416moles
Explanation:
Using the formula;
number of moles = mass ÷ molar mass
Molar mass of CO2 = 12 + 16(2)
= 12 + 32
= 44g/mol
According to the information in this question, there are 18.3g of CO2
number of moles (n) = 18.3/44
= 0.4159
= 0.416moles of CO2
Answer:
1.209g of MgO participates
Explanation:
In this problem, we have 0.030 moles of MgO that participates in a particular reaction.
And we are asked to solve for the mass of MgO that participates, that means, we need to convert moles to grams.
To convert moles to grams we need to use molar mass of the compound:
<em>1 atom of Mg has a molar mass of 24.3g/mol</em>
<em>1 atom of O has a molar mass of 16g/mol</em>
<em />
That means molar mass of MgO is 24.3g/mol + 16g/mol = 40.3g/mol
And mass of 0.030 moles of MgO is:
0.030 moles MgO * (40.3g/mol) =
<h3>1.209g of MgO participates</h3>