Answer:
Explanation:Artificial selection is distinct from natural selection in that it describes selection applied by humans in order to produce genetic change. When artificial selection is imposed, the trait or traits being selected are known, whereas with natural selection they have to be inferred. In most circumstances and unless otherwise qualified, directional selection is applied, i.e., only high-scoring individuals are favored for a quantitative trait. Artificial selection is the basic method of genetic improvement programs for crop plants or livestock (see Selective Breeding). It is also used as a tool in the laboratory to investigate the genetic properties of a trait in a species or population, for example, the magnitude of genetic variance or heritability, the possible duration of and limits to selection, and the correlations among traits, including with fitness.
Answer:
Percent yield = 94.5%
Theoretical yield = 26.89 g
Explanation:
Given data:
Mass of copper = 12.5 g
Mass of copper chloride produced = 25.4 g
Theoretical yield = ?
Percent yield = ?
Solution:
Cu + Cl₂ → CuCl₂
Number of moles of Copper:
Number of moles = mass/ molar mass
Number of moles = 12.5 g/ 63.55 g/mol
Number of moles = 0.2 mol
Now we will compare the moles of copper with copper chloride.
Cu : CuCl₂
1 : 1
0.2 : 0.2
Theoretical yield:
Mass of copper chloride:
Mass = Number of moles × molar mass
Mass = 0.2 mol × 134.45 g/mol
Mass = 26.89 g
Percent yield:
Percent yield = Actual yield / theoretical yield × 100
Percent yield = 25.4 g/26.89 g × 100
Percent yield = 94.5%
Surface of a sphere=4πr²
volume of a sphere=(4/3)πr³
Data:
surface area =432 m²
volume=864 m³
ratio of surface area to volume=surface area / volume
ratio=432 / 864 =0.5
Answer. The ratio would be 0.5
Answer:
(a) 0.699 kJ/K
(b) -0.671 kJ/K
(c) 0.028 kJ/K
Explanation:
The Refrigerant-134a flows into the evaporator as a saturated liquid-vapor mixture and flows out as a saturated vapor at a saturation pressure of 160 kPa and temperature of -15.64°C (estimated from the Saturated Refrigerant-134a Temperature Table).
(a) The entropy change of the refrigerant (ΔS
) = Q/T
Q = 180 kJ
T = -15.64 + 273.15 = 257.51 K
ΔS = Q/T = 180/257.51 = 0.699 kJ/K
(b) The entropy change (ΔS
) of the cooled space (ΔS) = -Q/T
Q = -180 kJ
T = -5 + 273.15 = 268.15 K
ΔS = Q/T = -180/268.15 = -0.671 kJ/K
(c) The total entropy change for this process (ΔS
) = ΔS + ΔS = 0.699 - 0.671 = 0.028 kJ/K
Hello
We know that polar molecules' attraction is stronger than that of non polar ones; this can be a good basis for us to check the substances' polarity. Knowing that polar forces are stronger, we can eliminate all options referring to them as being the weaker force; that is, options C and A. Next, we know that stronger forces will result in lower volatility so we can eliminate option D. The answer is B, assuming the compound has an odor.
