<span>In the 19th century, scientists realized that gases in the atmosphere cause a "greenhouse effect" which affects the planet's temperature. These scientists were interested chiefly in the possibility that a lower level of carbon dioxide gas might explain the ice ages of the distant past. At the turn of the century, Svante Arrhenius calculated that emissions from human industry might someday bring a global warming. Other scientists dismissed his idea as faulty. In 1938, G.S. Callendar argued that the level of carbon dioxide was climbing and raising global temperature, but most scientists found his arguments implausible. It was almost by chance that a few researchers in the 1950s discovered that global warming truly was possible. In the early 1960s, C.D. Keeling measured the level of carbon dioxide in the atmosphere: it was rising fast. Researchers began to take an interest, struggling to understand how the level of carbon dioxide had changed in the past, and how the level was influenced by chemical and biological forces. They found that the gas plays a crucial role in climate change, so that the rising level could gravely affect our future. (This essay covers only developments relating directly to carbon dioxide, with a separate essay for Other Greenhouse Gases. Theories are discussed in the essay on Simple Models of Climate.)</span>
Answer: Oxygen and glucose.
Explanation:
Oxygen and glucose are both reactants in the process of cellular respiration. The main product of cellular respiration is ATP; waste products include carbon dioxide and water.Jun 1, 2020
H₂CO₃ ⇔ HCO₃⁻ + H⁺
I 0.160 0 0
C -x +x +x
E 0.160-x +x +x
Ka1 = [HCO₃⁻][H⁺] / [H₂CO₃]
4.3 x 10⁻⁷ = x² / (0.160-x) (x is neglected in 0.160-x = 0.160)
x² = 6.88 x 10⁻⁸
x = 2.62 x 10⁻⁴
HCO₃⁻ ⇔ CO₃⁻² + H⁺
I 2.62 x 10⁻⁴ 0 2.62 x 10⁻⁴
C -x +x +x
E 2.62 x 10⁻⁴ - x +x 2.62 x 10⁻⁴ + x
Ka2 = [CO₃⁻²][H⁺] / [HCO₃⁻]
5.6 x 10⁻¹¹ = x(2.62 x 10⁻⁴ + x) / (2.62 x 10⁻⁴ - x)
x = 5.6 x 10⁻¹¹
Thus,
[H₂CO₃] = 0.160 - (2.62 x 10⁻⁴) = 0.16 M
[HCO₃⁻] = 2.62 x 10⁻⁴ - ( 5.6 x 10⁻¹¹) = 2.6 x 10⁻⁴ M
[CO₃⁻²] = 5.6 x 10⁻¹¹ M
[H₃O⁺] = 2.62 x 10⁻⁴ + 5.6 x 10⁻¹¹ = 2.6 x 10⁻⁴ M
[OH⁻] = 3.8 x 10⁻¹¹
Answer:
2Al+1.5O2→Al2O3
Thus, 2 mol of Al combine with 1.5 mol of oxygen to form 1 mol of Al2O3.
2 mol of Al corresponds to 2×27=54 g.
Thus, the weight of Al used in the reaction is 54 g.
Answer:
(a) C5H8O2
(b) C2H2Cl2
(c) CH2
(d) CH
Explanation:
We need to find the proportion of the atoms in whole numbers. Given the percentages we can calculate the number of moles and find their proportions.
Assume 100 g and given the atomic weights the moles are calculated.
(a) C = 59.9/ 12.01 = 4.98 ≈ 5.00
H = 8.06/1.007 = 8.00
O = 32/15.999 = 2.00
C5H8O2
(b) C= 24.8/12.01 = 2.06≈ 2.00
H = 2.0/1.007 = 1.99 ≈ 2.00
Cl = 73.1/ 35.453 = 2.06 ≈ 2.00
C2H2Cl2
(c) C = 86/12.01 = 7.16
H= 14/1.007 = 13.90
7.16:13.90 ≈ 1:2
CH2
(d) C = 92.30/12.01 = 7.68
H = 7.7 / 1.007 = 7.65
7.68:7.65 ≈ 1:1
CH
