Answer:
Horse latitude, trade winds
Explanation:
- The area of the low pressure or the calm consists of the variable light winds that blow near the equator are known to the marines as the doldrums and they form a circuital pattern near the earth atmosphere.
- Forms at a center of the near the higher pressure systems called as the horse latitudes where the trade winds at the surface are weak and variable and this zone is found generally in latitudes of the 30° North and South of the equator and move in an east to west direction.
Answer: Atoms with 11 protons, 10 neutrons and 11 electrons belong to the same element with 11 protons, 12 neutrons and 11 electrons.
Explanation:
Elements that contain same number of valence electrons belong to the same group. This is because they will have same reactivity (or properties) due to which they lie in the same group.
For example, element with 11 protons, 10 neutrons and 11 electrons is same as the element with 11 protons, 12 neutrons and 11 electrons.
Hence, both these atoms belong to the same element.
Thus, we can conclude that atoms with 11 protons, 10 neutrons and 11 electrons belong to the same element with 11 protons, 12 neutrons and 11 electrons.
With a radius of 3,760 miles (6,052 kilometers), Venus is roughly the same size as Earth — just slightly smaller.
Note: The question is incomplete. The complete question is given below :
Suppose a substance has a heat of fusion equal to 45 cal/g and a specific heat of 0.75 cal/g°C in the liquid state. If 5.0 kcal of heat are applied to a 50 g sample of the substance at a temperature of 24°C, what will its new temperate be? What state will the sample be in? (melting point of the substance = 27°C; specific heat of the solid =0.48 cal/g°C; boiling point of the substance = 700°C)
Explanation:
1.a) Heat energy required to raise the temperature of the substance to its melting point, H = mcΔT
Mass of solid sample = 50 g; specific heat of solid = 0.75 cal/g; ΔT = 27 - 24 = 3 °C
H = 50 × 0.75 × 3 = 112.5 calories
b) Heat energy required to convert the solid to liquid at its melting point at 27°C, H = m×l, where l = 45 cal/g
H = 50 × 45 = 2250 cal
c) Total energy used so far = 112.5 cal + 2250 cal = 2362.5 calories.
Amount of energy left = 5000 - 2362.5 = 2637.5 cal
The remaining energy is used to heat the liquid
H = mcΔT
Where specific heat of the liquid, c = 0.75 cal/g/°C, H = 2637.5 cal, ΔT = temperature change
2637.5 = 50 × 0.75 x ΔT
ΔT = 2637.5 / ( 50*0.75)
ΔT = 70.3 °C
Final temperature of sample = (70.3 + 27) °C = 97.3 °C
The substance will be in liquid state at a temperature of 97.3 °C
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