1. In a hydrogen fuel cell, hydrogen gas and oxygen gas are combined to form water. (1) Write the balanced chemical equation describing this reaction using the lowest whole number coefficients. (2) Identify the oxidizing agent and reducing agent. (3) Determine the number of electrons transferred in the balanced chemical equation
Answers
Explanation:
Hydrogen + Oxygen --> Water
(1) Write the balanced chemical equation describing this reaction using the lowest whole number coefficients.
2H2(g) + O2(g) ---> 2H2O(g)
(2) Identify the oxidizing agent and reducing agent.
Oxidizing agent = O (There is decrease in oxidation number from 0 to -2)
Reducing agent = H (There is increase in oxidation number form 0 to +1)
(3) Determine the number of electrons transferred in the balanced chemical equation
2H2(g) --> 4H+ + 4e- (4 hydrogen atom lost a single electron each)
O2 + 4e- --> 2O2- (Two oxygen gain 2 electrons each)
Total number of electrons transferred in the balanced chemical equation is 4.
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In their elemental forms, the halogens are Select one: a. strong oxidizing agents. b. strong reducing agents. c. strong acids. d. strong bases. e. amphoteric.
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Iron has density of 7.9g/cm3. What is the mass of a cube with the length of one side equal to 1.64x10squared cm3?
(B) Ca3PO6
(C) Ca4P2O4
(D) Ca3P2O8 (or Ca3(PO4)2)
(E) CaPO4
Answers
Answer:
D) empirical formula is: C₃P₂O₈
Explanation:
Given:
Mass % Calcium (Ca) = 38.7%
Mass % Phosphorus (P) = 19.9%
Mass % oxygen (O) = 41.2 %
This implies that for a 100 g sample of the unknown compound:
Mass Ca = 38.7 g
Mass P = 19.9 g
Mass O = 41.2 g
Step 1: Calculate the moles of Ca, P, O
Atomic mass Ca = 40.08 g/mol
Atomic mass P = 30.97 g/mol
Atomic mass O = 16.00 g/mol
Step 2: Calculate the molar ratio
Step 3: Calculate the closest whole number ratio
C: P: O = 1.50 : 1.00 : 4.00
C : P : O = 3:2:8
Therefore, the empirical formula is: C₃P₂O₈
Final answer:
The mass percentage composition of a compound can be used to determine its empirical formula. For a compound with 38.7% calcium (Ca), 19.9% phosphorus (P), and 41.2% oxygen (O), the empirical formula is Ca3(PO4)2.
Explanation:
To solve this problem, we're going to use the atomic mass percentages to determine the empirical formula of the compound.
We do this by assuming we have a 100g sample of the compound. Therefore:
The mass of calcium (Ca) is 38.7g.
The mass of phosphorus (P) is 19.9g.
The mass of oxygen (O) is 41.2g.
Next, we calculate how many moles we have of each element:
- Ca: 38.7g / 40.08g/mol (the atomic mass of calcium) = 0.965 moles
- P: 19.9g / 30.97g/mol (the atomic mass of phosphorous) = 0.643 moles
- O: 41.2g / 16.00g/mol (the atomic mass of oxygen) = 2.575 moles
Then, we divide each of these numbers by the smallest number of moles, which is 0.643 (P):
- Ca: 0.965/0.643 = 1.5 (~1)
- P: 0.643/0.643 = 1
- O: 2.575/0.643 = 4
Learn more about Empirical Formula here:
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Answers
Answer:
(c) CH₂F₂
Explanation:
Hydrogen bonds are weak intermolecular forces. They are the strongest kind of intermolecular forces, although they are weaker than the covalent bonds.
Hydrogen bonds arise from molecules which contain a hydrogen atom which is bonded to one of the most electronegative elements such as N, O or F.
(a) HF, → has H-F bond
(b) CH₃NH₂, → has N-H bond
(c) CH₂F₂, → has no H-F bond ( F- C- F)
(d) HOCH₂CH₂OH, → has O-H bond
Therefore, only CH₂F₂ does not exhibit hydrogen bonding.