Answer:
CH4 + 2 O2 → CO2 + 2 H2O
Explanation:
There are one mole of O2 on the left side and on the right side there are three moles of O2. And to fix it you would need to make it two moles of O2 to have four molecules of O2 on the left side. Then you would make two moles of H2O to have a total of four moles of O2 on the right. Therefore, CH4 + 2 O2 → CO2 + 2 H2O is the answer.
Answer:
See explanations
Explanation:
a. Molarity = moles/Volume in Liters = 5moles/2Liters = 2.5M in NaCl
b. Freezing Pt Depression
1. Sprinkling salt on icy surfaces
2. Using antifreeze in automobile cooling systems
3. <em>Not an application
</em>
4. Using salt to make ice cream
c. pOH = -log[OHˉ] = -log(1x10ˉ¹⁰) = -(-10) = 10 => pH = 14 – pOH = 14 – 10 = 4
d. H₂O + NH₃ => NH₄⁺ + OHˉ => Bronsted Acid is H₂O (proton donor)
Answer:
4
Explanation:
Ionization energy can be defined as the energy required for an atom to lose its valence electron to form an ion. Hence, it deals with how easily an atom would lose its electron and form an ion. As the valence electrons are lossless bound to the outermost shell, they can easily be lost without much problem or better still they can be lost easily. Hence, the energy change here is small and thus we can conclude that the ionization energy here is low.
The electron affinity works quite differently from the ionization energy. It deals with the way in which a neutral atom attracts an electron to form an ion. For an electron with loose valence electrons, the sure fact is that it does not really need these electrons. Hence, there is no need for an high electron affinity on its part. Thus, we conclude that the electron affinity is also low
Answer:
the compound contains C, H, and some other element of unknownidentity, so we can’t calculate the empirical formula
Explanation:
Mass of CO2 obtained = 3.14 g
Hence number of moles of CO2 = 3.14g/44.0 g = 0.0714 mol
The mass of the carbon in the sample = 0.0714 mol × 12.0g/mol = 0.857 g
Mass of H2O obtained = 1.29 g
Hence number of moles of H2O = 1.29g/18.0 g = 0.0717 mol
The mass of the carbon in the sample = 0.0717 mol × 1g/mol = 0.0717 g
% by mass of carbon = 0.857/1 ×100 = 85.7 %
% by mass of hydrogen = 0.0717/1 × 100 = 7.17%
Mass of carbon and hydrogen = 85.7 + 7.17 = 92.87 %
Hence, there must be an unidentified element that accounts for (100 - 92.87) = 7.13% of the compound.
