Answer:
27.7 grams of Li
0.140 moles of Co
0.158 moles of K
59.9 grams of As
8.00x10⁻⁹ moles of U
142.8 moles of H
0.0199 moles of O
0.666 moles of Pb
Explanation:
Mass / Molar mass = Mol
Mol . Molar mass = Mass
3.99 m . 6.94 g/m = 27.7 grams of Li
8.28 g / 58.93 g/m = 0.140 moles of Co
6.17 g / 39.1 g/m = 0.158 moles of K
0.8 mol . 74.92 g/m = 59.9 grams of As
6.02x10²³ (NA) particles are contained in 1 mol
6.02x10²³ are contained in 1 mol U____ 1 mol H ____ 1 mol O ___ 1 mol Pb
4.82x10¹⁵ atoms U ____ 4.82x10¹⁵/ 6.02x10²³ = 8.00x10⁻⁹ moles of U
8.60x10²⁵atoms H ____ 8.60x10²⁵ / 6.02x10²³ = 142.8 moles of H
1.20x10²² atoms O ____ 1.20x10²² / 6.02x10²³ = 0.0199 moles of O
4.01x10²³ atoms Pb ____ 4.01x10²³ / 6.02x10²³ = 0.666 moles of Pb
Kr 4d10 5s2 5p2 would be your answer.
Answer:
6 MOLES
YES MATES! TWO MOLES OF N2 REQIURS 6 MOLES OF H2 GASE TO REACT COMPLETLY
Answer:
B. They oxidize hydrocarbons to form less toxic gases.
Explanation:
A catalytic converter can be defined as an anti-pollution device containing a catalyst like platinum-iridium, installed in the exhaust chamber of an automobile so as to chemically convert harmful (poisonous) pollutants such as unburned hydrocarbons and carbon monoxide (CO), sulfur dioxide (S02), nitrogen oxide (NO) etc., into less harmful, poisonous or toxic chemical compounds.
This ultimately implies that, catalytic converters are typically used for converting harmful gases into less harmful, poisonous or toxic gases and molecules e.g carbon dioxide (C02) and water (H2O). This helps to prevent global warming, enhance the conservation of natural resources, as well as preserve the lives of living organisms and their natural habitat.
<em>Hence, the statement which best describes the use of catalytic converters in automobiles is that they oxidize hydrocarbons to form less toxic gases.</em>
Answer:
58 g/mol
Explanation:
According to Graham's law, the rate of diffusion of a gas (r) is inversely proportional to the square root of its molar mass (M). Butane's rate of diffusion is 3.8 times slower than that of helium, that is, rButane = rHe/3.8, or rHe/rButane = 3.8. Then,
