This problem handles<em> boiling-point elevation</em>, which means we will use the formula:
ΔT = Kb * m
Where ΔT is the difference of Temperature between boiling points of the solution and the pure solvent (Tsolution - Tsolvent). Kb is the ebullioscopic constant of the solvent (2.64 for benzene), and m is the molality of the solution.
Knowing that benzene's boiling point is 80.1°C, we <u>solve for m</u>:
Tsolution - Tsolvent = Kb * m
80.23 - 80.1 = 2.64 * m
m = 0.049 m
We use the definition of molality to <u>calculate the moles of azulene</u>:
0.049 m = Xmoles azulene / 0.099 kgBenzene
Xmoles azulene = 4.87 x10⁻³ moles azulene
We use the mass and the moles of azulene to<u> calculate its molecular weight</u>:
0.640 g / 4.875 x10⁻³ mol = 130.28 g/mol
<em>A molecular formula that would fulfill that molecular weight</em> is C₁₀H₁₀. So that's the result of solving this problem.
The actual molecular formula of azulene is C₁₀H₈.
Answer:
82.9 mL
Explanation:
1. Volume of silver
2. Volume of gold
3. Total volume of silver and gold
V = 4.766 mL + 2.591 mL = 7.36 mL
4 New reading of water level
V = 75.5 mL + 7.36 mL = 82.9 mL
<span>Electrons in a nitrogen-phosphorus covalent bond are not shared equally because nitrogen and phosphorus do not have the same electronegativity. The atoms spend more time around the most electronegative atom nitrogen.</span>
Answer:
Explanation:Since the compound X has no net-dipole moment so we can ascertain that this compound is not associated with any polarity.
hence the compound must be overall non-polar. The net dipole moment of compound is zero means that the vector sum of individual dipoles are zero and hence the two individual bond dipoles associated with C-Cl bond must be oriented in the opposite directions with respect to each other.]
So we can propose that compound X must be trans alkene as only in trans compounds the individual bond dipoles cancel each other.
If one isomer of the alkene is trans then the other two isomers may be cis .
Since the two alkenes give the same molecular formula on hydrogenation which means they are quite similar and only slightly different.
The two possibility of cis structures are possible:
in the first way it is possible the one carbon has two chlorine substituents and the carbon has two hydrogens.
Or the other way could be that two chlorine atoms are present on the two carbon atoms in cis manner that is on the same side and two hydrogens are also present on the different carbon atoms in the same manner.
Kindly refer the attachments for the structure of compounds:
Same as balancing a regular chemical reaction! Please see the related question to the bottom of this answer for how to balance a normal chemical reaction. This is for oxidation-reduction, or redox reactions ONLY! These instructions are for how to balance a reduction-oxidation, or redox reaction in aqueous solution, for both acidic and basic solution. Just follow these steps! I will illustrate each step with an example. The example will be the dissolution of copper(II) sulfide in aqueous nitric acid, shown in the following unbalanced reaction: CuS (s) + NO 3 - (aq) ---> Cu 2+ (aq) + SO 4 2- (aq) + NO (g) Step 1: Write two unbalanced half-reactions, one for the species that is being oxidized and its product, and one for the species that is reduced and its product. Here is the unbalanced half-reaction involving CuS: CuS (s) ---> Cu 2+ (aq) + SO 4 2- (aq) And the unbalanced half-reaction for NO 3 - is: NO 3 - (aq) --> NO (g) Step 2: Insert coefficients to make the numbers of atoms of all elements except oxygen and hydrogen equal on the two sides of each half-reaction. In this case, copper, sulfur, and nitrogen are already balanced in the two half-reaction, so this step is already done here. Step 3: Balance oxygen by adding H 2 O to one side of each half-reaction. CuS + 4 H 2 O ---> Cu 2+ + SO 4 2- NO 3 - --> NO + 2 H 2 O Step 4: Balance hydrogen atoms. This is done differently for acidic versus basic solutions. . For acidic solutions: Add H 3 O + to each side of each half-reaction that is "deficient" in hydrogen (the side that has fewer H's) and add an equal amount of H 2 O to the other side. For basic solutions: add H 2 O to the side of the half-reaction that is "deficient" in hydrogen and add an equal amount of OH - to the other side. Note that this step does not disrupt the oxygen balance from Step 3. In the example here, it is in acidic solution, and so we have: CuS + 12 H 2 O ---> Cu 2+ + SO 4 2- + 8 H 3 O + . NO 3 - + 4 H 3 O + --> NO + 6 H 2 O Step 5: Balance charge by inserting e - (electrons) as a reactant or product in each half-reaction. Oxidation: CuS + 12 H 2 O ---> Cu 2+ + SO 4 2- + 8 H 3 O + + 8 e - . Reduction: NO 3 - + 4 H 3 O + + 3 e - --> NO + 6 H 2 O . Step 6: Multiply the two half-reactions by numbers chosen to make the number of electrons given off by the oxidation step equal to the number taken up by the reduction step. Then add the two half-reactions. If done correctly, the electrons should cancel out (equal numbers on the reactant and product sides of the overall reaction). If H 3 O + , H 2 O, or OH - appears on both sides of the final equation, cancel out the duplication also. Here the oxidation half-reaction must be multiplied by 3 (so that 24 electrons are produced) and the reduction half-reaction must by multiplied by 8 (so that the same 24 electrons are consumed). 3 CuS + 36 H 2 O ---> 3 Cu 2+ + 3 SO 4 2- + 24 H 3 O + + 24 e - 8 NO 3 - + 32 H 3 O + + 24 e - ---> 8 NO + 48 H 2 O Adding these two together gives the following equation: 3 CuS + 36 H 2 O + 8 NO 3 - + 8 H 3 O + ---> 3 Cu 2+ + 3 SO 4 2- + 8 NO + 48 H 2 O Step 7: Finally balancing both sides for excess of H 2 O (On each side -36) This gives you the following overall balanced equation at last: 3 CuS (s) + 8 NO 3 - (aq) + 8 H 3 O + (aq) ---> 3 Cu 2+ (aq) + 3 SO 4 2- (aq) + 8 NO (g) + 12 H 2 O (l)
