can be oxidized to form carbon dioxide
Explanation:
Oxygen is important in the oxidation of glucose because it can be oxidized to form carbon dioxide. Oxidation of glucose involves the reaction of oxygen with glucose in a process called respiration. This gives a product of water, carbon dioxide and energy which is stored as ATP.
- Oxidation involves the addition of oxygen.
- Any specie that undergoes oxidation, is a reducing agent and it is said to be oxidized.
- Oxygen is oxidized to form carbon dioxide and water.
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I am pretty sure that the answer is B. extrusive rock. I hope that this helps you!! Good Luck!!
Explanation:
Compounds having same molecular formula but different structural and spatial arrangement are isomers.
Three isomers are possible for dibromomethene.
In one structure (IUPAC name: 1,1-dibromomethene), both the bromine atoms are attached to one carbon atom.
In another two structures (Cis and trans), two bromine atoms are attached to two different carbon atoms.
In Cis 1,2-dibromomethene, two bromine atoms are present on the same side.
Whereas in Cis 1,2-dibromomethene, two bromine atoms are present on the opposite side and hence, does not have net dipole moment.
Answer:
Electrolysis
Explanation:
The electrolysis of water is one such experiment that shows that water is made up of hydrogen and oxygen atoms only in the ratio of 2 to 1.
In the electrolysis of water, electricity is passed through acidified water to cause it to decompose.
The electrolysis of water is also known as the electrolysis of dilute tetraoxosulphate (VI) acid.
At the cathode, H⁺ ions are discharged and hydrogen gas is liberated:
2H⁺ + 2e⁻ → H₂
At the anode, both the sulfate ion and hydroxyl ions migrate to this electrode. Only the OH⁻ is selected for preferential discharge due to its lower position in that activity series.
4OH⁻ → 2H₂O + O₂ + 4e⁻
Oxygen gas is produced at the anode.
This electrolysis demonstrates the volumetric composition of water that is, 2 volumes of hydrogen at the cathode and 1 volume of oxygen at the anode.
You need the set of reactions that goes from ammonia to nitric acid.
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1) 4NH3(g)+5O2(g)-->4NO(g)+6H2O(g)
2) 2NO(g)+O2(g)-->2NO2(g)
3) 3NO2(g)+H2O(l)-->2HNO3(aq)+NO(g)
State the ratio of moles of HNO3 to NH3:
4 moles of NH3 produce 4 mole of NO,
4 moles of NO produce 4 moles of NO2
4 moles of NO2 produce 4 * (2 / 3) moles of HNO3 = 8/3 moles of HNO3.
=> (8/3) moles HNO3 : 4 moles NH3
Calculate the number of moles of HNO3 in 900.00 l of 0.140 M solution
M = n / V => n = M * V = 0.140 M * 900.00 liter = 126 moles HNO3
Use proportions:
(</span><span>8/3) moles HNO3 / 4 moles NH3 = 126 moles HNO3 / x
=> x = 126 moles HNO3 * 4 moles NH3 / (8/3 moles HNO3) = 189 moles NH3
Convert moles to grams:
molar mass NH3 = 14 g/mol + 3 * 1g/mol = 17 g/mol
mass in grams = number of moles * molar mass = 189 moles * 17 g/mol = 3213 g
Answer: 3213 g.
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