Answer:
carbon dioxide and water
Explanation:
Example: Combustion of Methane (CH₄(g))
CH₄(g) + 2O₂(g) => CO₂(g) + 2H₂O(g)**
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Note: The combustion of any hydrocarbon produces CO₂ & H₂O. That is,
Ethane (C₂H₆) + O₂ => CO₂(g) + H₂O(g)
Propane (C₃H₈) + O₂ => CO₂(g) + H₂O(g)
Butane (C₄H₁₀) + O₂ => CO₂(g) + H₂O(g)
The issue remaining is to balance the reaction equation. For these type equation balance Carbon 1st, then Hydrogen and finish with Oxygen. Balancing in this order leaves Oxygen which can be balanced using fractions. If problem requires lowest whole number ratios of elements, simply multiply entire equation by 2 to get standard equation*
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*Standard Equation is defined as the smallest whole number ratios of elements. The 'standard equation' is significant in that it is assumed to be at STP conditions; i.e., 0⁰C (=273K) & 1.0 Atmosphere pressure.
- Ethane (C₂H₆) + 7/2O₂(g) => 2CO₂(g) + 3H₂O(g)
=> 2C₂H₆ + 7O₂(g) => 4CO₂(g) + 6H₂O(g) <= Standard Form of Rxn
- Propane (C₃H₈) + 5O₂(g) => 3CO₂(g) + 4H₂O(g) <= Standard Form of Rxn (no need to balance with the '2' multiple)
- Butane (C₄H₁₀) + 13/2O₂ => 4CO₂(g) + 5H₂O(g)
=> 2C₃H₈ + 13O₂(g) => 4CO₂(g) + 5H₂O(g) <= Standard Form of Rxn
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**Also, note that water, H₂O(g), is listed as a gas. In some cases it will be listed as a liquid, H₂O(l).
Probably telophase because for the difference in the nuclear membrane
Answer:
65.08 g.
Explanation:
- For the reaction, the balanced equation is:
<em>2AlCl₃ + 3Br₂ → 2AlBr₃ + 3Cl₂,</em>
2.0 mole of AlCl₃ reacts with 3.0 mole of Br₂ to produce 2.0 mole of AlBr₃ and 3.0 mole of Cl₂.
- Firstly, we need to calculate the no. of moles of 36.2 grams of AlCl₃:
<em>n = mass/molar mass</em> = (36.2 g)/(133.34 g/mol) = <em>0.2715 mol.</em>
<u><em>Using cross multiplication:</em></u>
2.0 mole of AlCl₃ reacts with → 3.0 mole of Br₂, from the stichiometry.
0.2715 mol of AlCl₃ reacts with → ??? mole of Br₂.
∴ The no. of moles of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = (0.2715 mol)(3.0 mole)/(2.0 mole) = 0.4072 mol.
<em>∴ The mass of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = no. of moles of Br₂ x molar mass</em> = (0.4072 mol)(159.808 g/mol
) = <em>65.08 g.</em>
Answer: The amount of time needed to plate 14.0 kg of copper onto the cathode is 295 hours
Explanation:
We are given:
Moles of electron = 1 mole
According to mole concept:
1 mole of an atom contains 
number of particles.
We know that:
Charge on 1 electron = 
Charge on 1 mole of electrons = 
is passed to deposit = 1 mole of copper
63.5 g of copper is deposited by = 193000 C
of copper is deposited by =
To calculate the time required, we use the equation:
where,
I = current passed = 40.0 A
q = total charge = 42551181 C
t = time required = ?
Putting values in above equation, we get:
Converting this into hours, we use the conversion factor:
1 hr = 3600 seconds
So, 
Hence, the amount of time needed to plate 14.0 kg of copper onto the cathode is 295 hours
Nonmetals form negatively charged ions, or anions. They do this because they need to gain one to three electrons in order to achieve an octet of valence electrons, making them isoelectronic with the noble gas at the end of the period to which they belong.
