Calculate the percent composition for each of the elements in Na3PO4. A 5.00 gram sample of an oxide of lead PbxOy contains 4.33
g of lead. Determine the simplest formula for the compound. Balance the following equations by inspection: Cu(N03)2 rightarrow CuO + NO2 + O2 C2H6 + O2 rightarrow CO2 + H2O Mg3N2 + H2O rightarrow Mg(OH)2 + NH3 Magnesium (Mg) reacts with chlorine gas (C12) to produce magnesium chloride (MgC12). How many grams of Mg will react with 2 moles of Cl2? Mg(s) + C12(g) rightarrow MgC12(s) The compound P4S3 is used in matches. It reacts with oxygen to produce P4O10 and SO2. The unbalanced chemical equation is shown below, P4S3(s) + O2(g) rightarrow P4O10(s) + SO2(g) What mass of SO2 is produced from the combustion of 0.331 g P4S3 in excess O2? Nitric oxide is made from the oxidation of ammonia. What mass of nitric oxide can be made from the reaction of 8.00 g NH3 with 17.0 g O2? 4 NH3(g) + 5 O2(g) rightarrow 4 NO(g) + 6 H2O(g) To what volume should you dilute 50.0 mL of 12 M HNO3 solution to obtain a 0.100 M HNO3 solution?
1 answer:
Answer:
1. Percentage composition of: Na = 42%; P = 19.0%; O = 39%
2. Simplest formula of compound is PbO₂
3. (i) 2Cu(NO₃) ---> 2CuO + 2NO₂ + 3O₂
(ii) 2C₂H₆ + 7O₂ ---> 4CO₂ + 6H₂O
(iii) Mg₃N₂ + 6H₂O ---> 3Mg(OH)₂ + 2NH₃
4. 48 g of MG will react with 2 moles of Cl₂
5. 0.288 g of SO2 will be produced from the combustion of 0.331 g P₄S₃ in excess O₂
6. 12.8 g of nitric oxide can be produced from the reaction of 8.00 g NH₃ with 17.0 g O₂
7. The stock acid solution should be diluted to 6000 mL or 6.0 L
Explanation:
The full explanation is found in the attachments below
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Answer:
<h3>1)</h3>Structure One:
- N: -2
- C: 0
- O: +1
Structure Two:
- N: 0
- C: 0
- O: -1
Structure Three:
- N: -1
- C: 0
- O: 0.
Structure Number Two would likely be the most stable structure.
<h3>2)</h3>- All five C atoms: 0
- All six H atoms to C: 0
- N atom: +1.
The N atom is the one that is "likely" to be attracted to an anion. See explanation.
Explanation:
When calculating the formal charge for an atom, the assumption is that electrons in a chemical bond are shared equally between the two bonding atoms. The formula for the formal charge of an atom can be written as:
.
For example, for the N atom in structure one of the first question,
- N is in IUPAC group 15. There are 15 - 10 = 5 valence electrons on N.
- This N atom is connected to only 1 chemical bond.
- There are three pairs, or 6 electrons that aren't in a chemical bond.
The formal charge of this N atom will be .
Apply this rule to the other atoms. Note that a double bond counts as two bonds while a triple bond counts as three.
<h3>1)</h3>Structure One:
- N: -2
- C: 0
- O: +1
Structure Two:
- N: 0
- C: 0
- O: -1
Structure Three:
- N: -1
- C: 0
- O: 0.
In general, the formal charge on all atoms in a molecule or an ion shall be as close to zero as possible. That rules out Structure number one.
Additionally, if there is a negative charge on one of the atoms, that atom shall preferably be the most electronegative one in the entire molecule. O is more electronegative than N. Structure two will likely be favored over structure three.
<h3>2)</h3>Similarly,
- All five C atoms: 0
- All six H atoms to C: 0
- N atom: +1.
Assuming that electrons in a chemical bond are shared equally (which is likely not the case,) the nitrogen atom in this molecule will carry a positive charge. By that assumption, it would attract an anion.
Note that in reality this assumption seldom holds. In this ion, the N-H bond is highly polarized such that the partial positive charge is mostly located on the H atom bonded to the N atom. This example shows how the formal charge assumption might give misleading information. However, for the sake of this particular problem, the N atom is the one that is "likely" to be attracted to an anion.