Answer : The correct option is, (C) 2, 4 and 5.
Explanation :
Combustion reaction : It is a type of reaction in which a hydrocarbon react with an oxygen molecule to give carbon dioxide, water as a product.
For example : Methane react with oxygen to give carbon dioxide and water.
In the given list of chemical substances, 
are in oxide form. They can not be both reactant and product of a single combustion reaction.
In the given list, 
is the only hydrocarbon which shows a combustion reaction. That means react with 
to give 
and 
as a product.
The balanced combustion reaction of is,
Therefore, the correct answer is, (C) 2, 4, and 5.
Answer:
b) +2 and +3.
Explanation:
Hello,
In this case, given the molecular formulas:
And:
We can relate the subscripts with the oxidation states by knowing that they are crossed when the compound is formed, for that reason, we notice that oxygen oxidation state should be -2 for both cases and the oxidation state of X in the first formula must be +2 since both X and O has one as their subscript as they were simplified:
Moreover, for the second case the oxidation state of X should be +3 in order to obtain 3 as the subscript of oxygen:
Thus, answer is b)+2 and +3
Best regards.
diatomic hydrogen is written as H2 (2.02 grams H2) <------- if each hydrogen atom is 1.01 grams, then two hydrogen atoms are 2.02 grams 2.0 moles H2 X 2.02 grams H2 ------------- (divide to cancel moles) = 4.04 grams/mole H2 ÷ one mole = 4.04 grams H2
Answer:
2--->C
6---->E
3---->D
4--->A
5--->B
1---->F
Explanation:
I think so, sorry if its wrong.
Answer:
Percent yield: 78.2%
Explanation:
Based on the reaction:
4Al + 3O₂ → 2Al₂O₃
<em>4 moles of Al produce 2 moles of Al₂O₃</em>
<em />
To find percent yield we need to find theoretical yield (Assuming a yield of 100%) and using:
(Actual yield (6.8g) / Theoretical yield) × 100
Moles of 4.6g of Al (Molar mass: 26.98g/mol) are:
4.6g Al × (1mol / 26.98g) = 0.1705 moles of Al.
As 4 moles of Al produce 2 moles of Al₂O₃, theoretical moles of Al₂O₃ obtained from 0.1705 moles of Al are:
0.17505 moles Al × (2 moles Al₂O₃ / 4 moles Al) = <em>0.0852 moles of Al₂O₃</em>,
In grams (Molar mass Al₂O₃ = 101.96g/mol):
0.0852 moles of Al₂O₃ × (101.96g / mol) =
<h3>8.7g of Al₂O₃ can be produced (Theoretical yield)</h3>
Thus, Percent yield is:
(6.8g / 8.7g) × 100 =
<h3>
78.2% </h3>
