Answer:
80 liters
Explanation:
At STP, 1 mole of ideal gas has a volume of 22.4 liters.
Therefore, since liters and moles are directly proportional, we can use stoichiometry directly.
40L CH₄ × (2L H₂ / 1L CH₄) = 80L H₂
Answer:
The percentage composition of each element in H2O2 is 5.88% H and 94.12% O (Option D).
Explanation:
Step 1: Data given
Molar mass of H = 1.0 g/mol
Molar mass of O = 16 g/mol
Molar mass of H2O2 = 2*1.0 + 2*16 = 34.0 g/mol
Step 2: Calculate % hydrogen
% Hydrogen = ((2*1.0) / 34.0) * 100 %
% hydrogen = 5.88 %
Step 3: Calculate % oxygen
% Oxygen = ((2*16)/34)
% oxygen = 94.12 %
We can control this by the following equation
100 % - 5.88 % = 94.12 %
The percentage composition of each element in H2O2 is 5.88% H and 94.12% O (Option D).
Answer:
B. Ionic Compound
Explanation:
An ionic compound is that compound which contains a positively charged ion called CATION and a negatively charged ion called ANION. The cation loses or transfers electrons to the anion, hence, making the former (cation) positive and the latter (anion) negative.
A polyatomic ion is an ion that contains more than one type of atom e.g OH-, NO3²-, CO3²- etc. A polyatomic ion usually has an overall charge formed from the charges of the individual atoms that makes it up. For example, in OH-, the overall charge is -1.
Since a polyatomic ion can have an overall positive or negative charge, it must enter a reaction with another ion that complements it i.e. a negative polyatomic ion will react with a positive ion to neutralize its charge. Hence, this forms an IONIC COMPOUND. This is why most compounds with polyatomic ions are IONIC COMPOUNDS.
For example, CaCO3 is an ionic compound formed when Ca²+ (cation) reacts with the polyatomic anion: CO3²-
<span>A compound is found to be 40.0% carbon, 6.7% hydrogen and 53.5% oxygen. Its molecular mass is 60. g/mol.
</span>Q1)
Empirical formula is the simplest ratio of whole numbers of components making up a compound.
the percentages have been given, therefore we can calculate for 100 g of the compound.
C H O
Mass in 100 g 40.0 g 6.7 g 53.5 g
Molar mass 12 g/mol 1 g/mol 16 g/mol
Number of moles 40.0/12= 3.33 6.7/1 = 6.7 53.5/16 = 3.34
Divide by the least number of moles
3.33/3.33 = 1 6.7/3.33 = 2.01 3.34/3.33 = 1.00
after rounding off
C - 1
H - 2
O - 1
Empirical formula - CH₂O
Q2)
Molecular formula is the actual number of components making up the compound.
To find the number of empirical units we have to find the mass of one empirical unit.
Mass of one empirical unit = CH₂O - 12 + (1x2) + 16 = 30 g
Mass of one mole of compound = 60 g
Number of empirical units = 60 g / 30 g = 2
Therefore molecular formula - 2(CH₂O)
Molecular formula - C₂H₄O₂
Answer:
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