Answer:
Define a problem, form a hypothesis, gather experimental data, form a conclusion
Answer:
29.3 g HClO₄
Explanation:
We have 39.1 grams of 74.9 wt% aqueous perchloric acid solution, that is, there are 74.9 grams of perchloric acid in 100 grams of perchloric acid solution. The mass, in grams, of perchloric acid contained in 39.1 grams of perchloric acid solution is:
39.1 g Solution × (74.9 g HClO₄/100 g Solution) = 29.3 g HClO₄
4. describe three ways carbon dioxide was removed from the Earth's atmosphere.
Answer: Forests: Photosynthisis helps clear carbon dioxide naturally, Soils naturally store carbon, but agricultural soils are running a big deficit due to intensive use. Because agricultural land is so expansive, Bio-energy with Carbon Capture and Storage (BECCS) is another way to use photosynthesis to combat climate change. However, it is far more complicated than planting trees or managing soils — and it doesn’t always work for the climate.
5. Explain why there is now 21% Oxygen in the Earth's atomosphere compaired to little or no Oxygen in the Earth's atmosphere 4.5 billion years ago.
Answer: cientists believe that the Earth was formed about 4.5 billion years ago. Its early atmosphere was probably formed from the gases given out by volcanoes. It is believed that there was intense volcanic activity for the first billion years of the Earth's existence.The early atmosphere was probably mostly carbon dioxide, with little or no oxygen. There were smaller proportions of water vapour, ammonia and methane. As the Earth cooled down, most of the water vapour condensed and formed the oceans.
Sorry its soooo long TwT
Answer:
t = 7.58 * 10¹⁹ seconds
Explanation:
First order rate constant is given as,
k = (2.303
/t) log [A₀]
/[Aₙ]
where [A₀] is the initial concentraion of the reactant; [Aₙ] is the concentration of the reactant at time, <em>t</em>
[A₀] = 615 calories;
[Aₙ] = 615 - 480 = 135 calories
k = 2.00 * 10⁻²⁰ sec⁻¹
substituting the values in the equation of the rate constant;
2.00 * 10⁻²⁰ sec⁻¹ = (2.303/t) log (615/135)
(2.00 * 10⁻²⁰ sec⁻¹) / log (615/135) = (2.303/t)
t = 2.303 / 3.037 * 10⁻²⁰
t = 7.58 * 10¹⁹ seconds
Answer:
There are 1.4754246675000002e+24 atoms of Hydrogen within the measurement of 2.45 moles of hydrogen!
Explanation:
