The chemical formula for the compound can be written as,
CxHyOz
where x is the number of C atoms, y is the number of H atoms, and z is the number of O atoms. The combustion reaction for this compound is,
CxHyOz + O2 --> CO2 + H2O
number of moles of C:
(0.7191 g)(1 mol CO2/44 g of CO2) = 0.0163 mol CO2
This signifies that 0.0163 mole of C and the mass of carbon in the compound,
(0.0163 mols C)(12 g C/ 1 mol C) = 0.196 g C
number of moles H:
(0.1472 g H2O)(1 mol H2O/18 g H2O) = 0.00818 mol H2O
This signifies that there are 0.01635 atoms of H in the compound.
mass of H in the compound = (0.01635 mols H)(1 g of H) = 0.01635 g H
Mass of oxygen in the compound,
0.3870 - (0.196 g C + 0.01635 g H) = 0.1746 g
Moles O in the compound = (0.1746 g O)(1 mol O/16 g O) = 0.0109 mols O
The formula of the compound is,
C0.0163H0.01635O0.0109
Dividing the numbers by the least number,
C3/2H3/2O
The empirical formula of the compound is therefore,
<em> C₃H₃O₂</em>
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)
Answer:
The correct answer is 0.300 * 10^23 ions.
Explanation:
Based on the given question, there is a need to find the number of chloride ions in the mentioned 6.8 grams of zinc chloride compound.
The moles of zinc chloride (ZnCl2) is,
= mass of zinc + 2 mass of chlorine
= 65.38 + 2 (35.45)
=65.38 + 70.90
= 136.28 grams (The molecular mass of zinc is 65.38 and the molecular mass of chlorine is 35.45)
Thus, 136.28 g of ZnCl2 contains 70.90 grams of chlorine
Therefore, 6.8 grams of ZnCl2 will comprise = (70.90/136.28) * 6.8
= 3.537 g of chlorine
70.90 g of Cl comprise 6.022*10^23 chlorine, thus, 3.537 g of Cl will comprise (6.022*10^23/70.90) * 3.537
= 0.300 * 10^23 ions of chlorine.
Answer:
the change is evaporation
Explanation:
the water heats up at the surface of the water and evaporates
