<span>0.0292 moles of sucrose are available.
First, lookup the atomic weights of all involved elements
Atomic weight Carbon = 12.0107
Atomic weight Hydrogen = 1.00794
Atomic weight Oxygen = 15.999
Now calculate the molar mass of sucrose
12 * 12.0107 + 22 * 1.00794 + 11 * 15.999 = 342.29208 g/mol
Divide the mass of sucrose by its molar mass
10.0 g / 342.29208 g/mol = 0.029214816 mol
Finally, round the result to 3 significant figures, giving
0.0292 moles</span>
The purport of mitosis is cell regeneration, magnification, and asexual reproduction,while the purport of meiosis is the engenderment of gametes for sexual reproduction. Mitosis is a single nuclear division that results in two nuclei that are conventionally partitioned into two incipient daughter cells.
Answer:
Depth and location affect ocean water’s temperature.
Explanation:
The main source of heat for the oceans is solar radiation. That is, water is basically heated by the radiation of the Sun, which transmits energy to the surface. The ocean absorbs this energy and stores it. Seawater has high caloric capacity. This means that more energy and more time is needed to change or increase the water temperature, compared to the air temperature. Similarly, once the ocean heats up, it takes a long time for the water to completely release or lose that heat.
The temperature decreases to greater depth, because the amount of solar radiation is reduced. On the contrary, it is greater where there is greater energy or heat content.
The closer a place is to the equator, the solar energy will affect more vertically and with more intensity on it, so the warmer the temperatures will be. The further that point of the equator is found, the solar energy will reach it with a smaller angle. And if the point is near the poles, the sun's rays will arrive at a very small angle. This causes the temperature of the water of the oceans to vary depending on the earth's latitude, being higher in areas close to the equator and the tropics, and colder the closer to the poles or the further away from the temperate zones.
Answer:
3.2 L
Explanation:
Given data:
Mass of oxygen = 3.760 g
Pressure of gas = 88.4 Kpa (88.4×1000 = 88400 Nm⁻²)
Temperature = 19°C (19+273.15 = 292.15 K)
R = 8.314 Nm K⁻¹ mol⁻¹
Volume occupied = ?
Solution:
Number of moles of oxygen:
Number of moles = mass/ molar mass
Number of moles = 3.760 g/ 32 g/mol
Number of moles = 0.12 mol
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant
T = temperature in kelvin
V = nRT/P
V = 0.12 mol × 8.314 Nm K⁻¹ mol⁻¹ × 292.15 K /88400 Nm⁻²
V = 291.472 Nm /88400 Nm⁻²
V = 0.0032 m³
m³ to L:
V = 0.0032×1000 = 3.2 L