If an object has density of 0.75 g/milliliters then which is greater the mass or the volume?
1 answer:
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Answer: 2 lone pairs, square planar
Explanation:
Using the VSEPR ( Valence Shell Electron Pair Repulsion)Theory
To calculate the number of lone pairs electron can be done using the formula;
Number of electrons = ½ (V+N-C+A)
V mean valency of the central atom
N means number of monovalent bonding atoms
C means charge on cation
A means charges on anion
Therefore, to calculate the number of lone pair electron C=A=0;
Number of electrons = ½ (8+4) = 12/2 = 6
Number of bonding pair = 4
Number of lone pairs of electron = 6-4 = 2
The hybridrization of the compound is sp3d2 because the number of electrons around the central atom is 6.
The geometry of the compound is square planar and this is because of the repulsion between the bonding pair of electrons and lone pair of electrons which causes the lone pair of electrons to lie in a perpendicular plane in order to acquire stability.
N2O4(l) + 2 N2H4(l) → 3 N2(g) + 4 H2O(g)
1) to calculate the limiting reactant you need to pass grams to moles.
<span> moles is calculated by dividing mass by molar mass
</span>
mass of N2O4: 50.0 g
molar mass of <span>N2O4 = 92.02 g/mol
</span><span>molar mass of N2H4 = 32.05 g/mol.
</span>mass of N2H4:45.0 g
moles N2O4=50.0/92.02 g/mol= 0,54 mol of N2O4
moles N2H4= 45/32.05 g/mol= 1,40 mol of <span><span>N2H4
</span> 2)</span>
By looking at the balanced equation, you can see that 1 mol of N2O4 needs 2 moles of N2H4 to fully react . So to react 0,54 moles of N2O4, you need 2x0,54 moles of <span>N2H4 moles
</span><span>N2H4 needed = 1,08 moles.
You have more that 1,08 moles </span><span>N2H4, so this means the limiting reagent is not N2H4, it's </span>N2O4. The molecule that has molecules that are left is never the limiting reactant.
3) 1 mol of N2O4 reacting, will produce 3 mol of N2 (look at the equation)
There are 0,54 mol of N2O4 available to react, so how many moles will produce of N2?
1 mol N2O4------------3 mol of N2
0,54 mol N2O4--------x
x=1,62 mol of N2
4) the only thing left to do is convert the moles obtained, to grams.
We use the same formula as before, moles equal to mass divided by molar mass.
(molar mass of N2= 28)
1,62 mol of N2= mass/ 28
mass of N2= 45,36 grams
1) to calculate the limiting reactant you need to pass grams to moles.
<span> moles is calculated by dividing mass by molar mass
</span>
mass of N2O4: 50.0 g
molar mass of <span>N2O4 = 92.02 g/mol
</span><span>molar mass of N2H4 = 32.05 g/mol.
</span>mass of N2H4:45.0 g
moles N2O4=50.0/92.02 g/mol= 0,54 mol of N2O4
moles N2H4= 45/32.05 g/mol= 1,40 mol of <span><span>N2H4
</span> 2)</span>
By looking at the balanced equation, you can see that 1 mol of N2O4 needs 2 moles of N2H4 to fully react . So to react 0,54 moles of N2O4, you need 2x0,54 moles of <span>N2H4 moles
</span><span>N2H4 needed = 1,08 moles.
You have more that 1,08 moles </span><span>N2H4, so this means the limiting reagent is not N2H4, it's </span>N2O4. The molecule that has molecules that are left is never the limiting reactant.
3) 1 mol of N2O4 reacting, will produce 3 mol of N2 (look at the equation)
There are 0,54 mol of N2O4 available to react, so how many moles will produce of N2?
1 mol N2O4------------3 mol of N2
0,54 mol N2O4--------x
x=1,62 mol of N2
4) the only thing left to do is convert the moles obtained, to grams.
We use the same formula as before, moles equal to mass divided by molar mass.
1,62 mol of N2= mass/ 28
mass of N2= 45,36 grams
Answer:
The boiling point is 308.27 K (35.27°C)
Explanation:
The chemical reaction for the boiling of titanium tetrachloride is shown below:
Ti ⇒ Ti
ΔH° (Ti
) = -804.2 kJ/mol
ΔH° (Ti
) = -763.2 kJ/mol
Therefore,
ΔH° = ΔH°
(Ti
) - ΔH°
(Ti
) = -763.2 - (-804.2) = 41 kJ/mol = 41000 J/mol
Similarly,
s°(Ti) = 221.9 J/(mol*K)
s°(Ti) = 354.9 J/(mol*K)
Therefore,
s° = s° (Ti) - s°(Ti
) = 354.9 - 221.9 = 133 J/(mol*K)
Thus, T = ΔH° /s° = [41000 J/mol]/[133 J/(mol*K)] = 308. 27 K or 35.27°C
Therefore, the boiling point of titanium tetrachloride is 308.27 K or 35.27°C.