Answer:
Quantity A is weight and Quantity B is mass
Explanation: weight has same unit as force. Mass is the quantity of matter present in a body or object
Answer:
C - commas indicate a natural pause in the sentence
Answer:
Explanation:
These instrument works on the analysis of the emisson spectral of light received from the star in this way.
Think of a steel knife in your kitchen. Initially, it has this shiny silver colour that typifies it. When the knife is placed on a hot plate, it becomes hotter and begins to go red as the heating continues. If we stop the heating and pour cold water on it, the red dissapears and our knife is back to itself, although the silvery shine would be lost. This is simply how the atomic absorption spectroscopy works. When you see the hot knife you can say a couple of things about it. Different metals have their various melting point. We can compare the temperature at which our knife will melt with a standard melting point scale to know the type of metal it is made of.
In atomic absorption spectroscopy, an atom gains energy and it becomes excited. Every atom is known to have a peculair amount of absorbant energy that cause them to excite. The more the particles in the atom, the more the energy required. When we analyse the absorbent energy of the atom, it differs from other atoms and we truly identify such an atom even if we don't know it. Most times, the energy is given off as light.
a. the ratio of mass to charge of an electron
Explanation:
The experiment permitted the direct measurement of the ratio of mass to charge of an electron.
- The charge to mass ratio of an electron was determine by accelerating a beam of cathode rays in magnetic and electric fields.
- No matter the gas used in the tube or the nature of the material of the electrodes, the rays were found to have constant charge to mass ratio of 1.76 x 10¹¹coulombkg⁻¹.
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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
i hope that this eg gonna help u
