Answers
Answer: The mass of carbon and hydrogen in the sample is 0.1087 g and 0.0066 g respectively and the percentage composition of carbon and hydrogen in the sample is 94.27 % and 5.72 % respectively.
Explanation:
The chemical equation for the combustion of hydrocarbon having carbon and hydrogen follows:
where, 'x' and 'y' are the subscripts of carbon and hydrogen respectively.
We are given:
Mass of
Mass of
Mass of sample = 0.1153 g
We know that:
Molar mass of carbon dioxide = 44 g/mol
Molar mass of water = 18 g/mol
- For calculating the mass of carbon:
In 44g of carbon dioxide, 12 g of carbon is contained.
So, in 0.3986 g of carbon dioxide, of carbon will be contained.
- For calculating the mass of hydrogen:
In 18g of water, 2 g of hydrogen is contained.
So, in 0.0578 g of water, of hydrogen will be contained.
To calculate the percentage composition of a substance in sample, we use the equation:
......(1)
- For Carbon:
Mass of sample = 0.1153 g
Mass of carbon = 0.1087 g
Putting values in equation 1, we get:
- For Hydrogen:
Mass of sample = 0.1153 g
Mass of hydrogen = 0.0066 g
Putting values in equation 1, we get:
Hence, the mass of carbon and hydrogen in the sample is 0.1087 g and 0.0066 g respectively and the percentage composition of carbon and hydrogen in the sample is 94.27 % and 5.72 % respectively.
Related Questions
Which of the following is a chemical property?A. Gallium metal melts in your hand.B. Platinum does not react with oxygen at room temperature.C. The page of the textbook is white.D. The copper sheets that form the "skin" of the Statue of Libertyhave acquired a greenish coating over the years.
Is 3.5:Aqueous solutions of iron(III) bromide and ammonium carbonate react to form a precipitate. Answer the follwingquestions with regards to this reaction.a) Write the molecular equation for this reaction byTranslating the two reactants into their chemical formulae.Predict the products.Label all the states.Balance the reaction.
What is planck's theory?
Animal fats tend to be _______________ with hydrogen atoms. (fifth paragraph of this sub-section
O neutrons
O Neither... An atom isn't neutral!
Answers
Taking into account the constitution of an atom, an atom is neutral because it has the same number of protons as it has electrons.
Constitution of an atom
An atom is the smallest constituent unit of ordinary matter that has the properties of a chemical element.
Every atom consists of a nucleus in which neutrons and protons meet and energy levels where electrons are located.
The neutron is an electrically neutral subatomic particle, while the proton has a positive electrical charge. Electrons have a negative charge, move around the nucleus at different energy levels and are attracted to protons, positive in the atom through electromagnetic force.
Atomelectrically neutral
An atom is considered electrically neutral when it has the same number of positive and negative charges. That is, an electrically neutral atom has the same number of protons (with a positive charge) and electrons (with a negative charge).
In summary, an atom is neutral because it has the same number of protons as it has electrons.
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Answers
Answer:
Polarity
Cohesion
Adhesion
High Specific Heat
Explanation:
Answers
Answer:
6.82 kg
Explanation:
Given that the amount of water is 15L and we know that the density of water is ≈ 1kg/L. The mass of water is given by mass = volume x density, i.e,
mass = 15 x 1 = 15 kg. Also the specific heat capacity of water is 4.186 KJ/kg.
The sublimation enthalpy of dry ice is 571 KJ/kg.
Now, the amount of heat lost by water is entirely used up for the sublimation (conversion from soild to gas) of dry ice. And the heat (Q) lost by water is given as : Q = mCΔT, where m is the mass of water, C the specific heat capacity of water and ΔT the change in temperature.
Here, Q = 15 x 4.186 x (90 - 28) = 3892.98 KJ.
This amount of heat is taken up by the dry ice for its sublimation. Also the energy taken by dry ice (Q') for its sublimation is given by: Q' = m'L', where m' is the mass of dry ice, L' is the latent heat of sublimation (i.e, the amount of heat required per kg of a substance to sublime) of dry ice amd L' = 571 KJ/kg.
Now, Q' =m'L' = heat lost by water = 3892.98KJ.
And, m'L' = m' x 571 KJ/kg = 3892.98 KJ. (Dividing with 571)
Therefore, m' = 6.82 kg.
Answers
Answer:
MnO- Manganese Oxide
Explanation:
Empirical formula: This is the formula that shows the ratio of elements
present in a
compound.
How to determine Empirical formula
1. First arrange the symbols of the elements present in the compound
alphabetically to determine the real empirical formula. Although, there
are exceptions to this rule, E.g H2So4
2. Divide the percentage composition by the mass number.
3. Then divide through by the smallest number.
4. The resulting answer is the ratio attached to the elements present in
a compound.
Mn O
% composition 72.1 27.9
Divide by mass number 54.94 16
1.31 1.74
Divide by the smallest number 1.31 1.31
1 1.3
The resulting ratio is 1:1
Hence the Empirical formula is MnO, Manganese oxide
Answers
0.0340 g O2
Step 1. Write the balanced chemical equation
4Fe(OH)^(+) + 4OH^(-) + O2 + 2H2O → 4Fe(OH)3
Step 2. Calculate the moles of Fe^(2+)
Moles of Fe^(2+) = 50.0 mL Fe^(2+) × [0.0850 mmol Fe^(2+)/1 mL Fe^(2+)]
= 4.250 mmol Fe^(2+)
Step 3. Calculate the moles of O2
Moles of O2 = 4.250 mmol Fe^(2+) × [1 mmol O2/4 mmol Fe^(2+)]
= 1.062 mmol O2
Step 4. Calculate the mass of O2
Mass of O2 = 1.062 mmol O2 × (32.00 mg O2/1 mmol O2) = 34.0 mg O2
= 0.0340 g O2
0.0342 grams of O2 are consumed to precipitate all of the iron in 50.0 mL of 0.0850 M Fe(II) solution.
To solve this problem, we need to first calculate the number of moles of Fe(II) in 50.0 mL of 0.0850 M Fe(II) solution.
Moles of Fe(II) = (0.0850 mol/L) * (50.0 mL) = 0.00425 mol
According to the balanced chemical equation, 4 moles of Fe(II) react with 1 mole of O2. Therefore, the number of moles of O2 required to precipitate all of the iron in 50.0 mL of 0.0850 M Fe(II) solution is:
Moles of O2 = (0.00425 mol Fe(II)) * (1 mol O2 / 4 mol Fe(II)) = 0.00106 mol O2
Now we can convert the moles of O2 to grams using the molar mass of O2 (32.00 g/mol):
Grams of O2 = (0.00106 mol O2) * (32.00 g/mol) = 0.0342 g O2
Therefore, 0.0342 grams of O2 are consumed to precipitate all of the iron in 50.0 mL of 0.0850 M Fe(II) solution.
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