The nitrogen atom in NH3 is sp3 hybridized. ... In NH3 molecule, three of the sp3 orbitals are used to formbonds to the three hydrogen atoms and the fourth sp3 orbital is used to hold the lone pair. and is sometimes called a non-bonding pair.
Answer: the product is ketone or aldehyde
Explanation:
The first step is the conversion of acetal to hemiacetal in the presence of H3O+/ ROH, and then the final conversion of hemiacetal to ketone/aldehyde using
H3O+/ ROH...
Attached is the structural conversion
Answer:
35,000,000,000 mL
Explanation:
You first multiply 35 times 1000.
35,000 L
Now you multiply 35,000 times 10^6
35,000,000,000 mL
In a combustion of a hydrocarbon compound, 2 reactions are happening per element:
C + O₂ → CO₂
2 H + 1/2 O₂ → H₂O
Thus, we can determine the amount of C and H from the masses of CO₂ and H₂O produced, respectively.
1.) Compute for the amount of C in the compound. The data you need to know are the following:
Molar mass of C = 12 g/mol
Molar mass of CO₂ = 44 g/mol
Solution:
0.5008 g CO₂*(1 mol CO₂/ 44 g)*(1 mol C/1 mol CO₂) = 0.01138 mol C
0.01138 mol C*(12 g/mol) = 0.13658 g C
Compute for the amount of H in the compound. The data you need to know are the following:
Molar mass of H = 1 g/mol
Molar mass of H₂O = 18 g/mol
Solution:
0.1282 g H₂O*(1 mol H₂O/ 18 g)*(2 mol H/1 mol H₂O) = 0.014244 mol H
0.014244 mol H*(1 g/mol) = 0.014244 g H
The percent composition of pure hydrocarbon would be:
Percent composition = (Mass of C + Mass of H)/(Mass of sample) * 100
Percent composition = (0.13658 g + 0.014244 g)/(<span>0.1510 g) * 100
</span>Percent composition = 99.88%
2. The empirical formula is determined by finding the ratio of the elements. From #1, the amounts of moles is:
Amount of C = 0.01138 mol
Amount of H = 0.014244 mol
Divide the least number between the two to each of their individual amounts:
C = 0.01138/0.01138 = 1
H = 0.014244/0.01138 = 1.25
The ratio should be a whole number. So, you multiple 4 to each of the ratios:
C = 1*4 = 4
H = 1.25*4 = 5
Thus, the empirical formula of the hydrocarbon is C₄H₅.
3. The molar mass of the empirical formula is
Molar mass = 4(12 g/mol) + 5(1 g/mol) = 53 g/mol
Divide this from the given molecular weight of 106 g/mol
106 g/mol / 53 g/mol = 2
Thus, you need to multiply 2 to the subscripts of the empirical formula.
Molecular Formula = C₈H₁₀
Answer:
3.72 mol Hg
General Formulas and Concepts:
<u>Chemistry - Atomic Structure</u>
- Reading a Periodic Table
- Using Dimensional Analysis
- Density = Mass over Volume
Explanation:
<u>Step 1: Define</u>
D = 13.6 g/mL
54.8 mL Hg
<u>Step 2: Identify Conversions</u>
Molar Mass of Hg - 200.59 g/mol
<u>Step 3: Find</u>
13.6 g/mL = x g / 54.8 mL
x = 745.28 g Hg
<u>Step 4: Convert</u>
<u />
= 3.71544 mol Hg
<u>Step 5: Check</u>
<em>We are given 3 sig figs. Follow sig fig rules and round.</em>
3.71544 mol Hg ≈ 3.72 mol Hg
