In the Early 1800s, John Dalton stated that matter is tiny particles called atoms that are indivisible. Some materials were only made of one kind of atom called an element. Two atoms of hydrogen combined with one atom of oxygen would form one molecule of water.
Then, in 1897, J.J Thompson proposed that repetitively charged particles inside of the atom were part of every atom. And those were called electrons. They called Thomson's work "Plum Pudding." Sadly, "Plum Pudding" didn't last very long.
Rutherford concluded there was a center of a positive charge within the atom called protons. Atoms contained one proton for each electron.
P.rotons- Positively electric charged
E.lectrons- Negative electricity
N.eutrons- about the same mass as a proton, but with no electric charge.
P.E.N
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The potassium metal atoms occupying the crystal lattice that occurs is being held together by metallic boding
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
Of what type of solutions
like any type
Answer:
The vapor pressure of ethanol in the solution is 10,27 kPa
Explanation:
To obtain the vapor pressure of a solution it is necessary to use Raoult's law:
<em>(1)</em>
The moles of ethanol are:
18,00mL×× = 0,3083 mol Ethanol.
Moles of benzoic acid:
12,55 g× = 0,1028 mol benzoic acid.
Thus, mole fraction of solvent, X, is:
=<em> 0,7499</em>
Replacing this value in (1):
= <em>10,27 kPa</em>
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I hope it helps!