Answer:
SAMPLE A - pure substance.
SAMPLE B - homogeneous mixture.
SAMPLE C - heterogeneous mixture.
Explanation:
Dalton Found out there was a small, hard indestructible sphere that is the smalles part of an element.He created his own Atomic Theory:
-All Matter is made up of small particles called atoms.
-Atoms cannot be created, destroyed, or divided into smaller particles.
-All atoms of the same element are identical in mass and size. The atoms of one element are different in mass and size from the atoms of other elements.
<span>-Compounds are created when atoms of different elements link together in definite proportions.
</span><span>Rutherford had found the positively charged nucleus in the middle of every atom using his Gold Foil Experiment. While doing this experiment, he expected these particles to just pass right through the foil but they bounced right back. He also proposed there were negatively charged electrons revolving around the nucleus.
</span><span>Thompson found negative electrons and inferred atoms also contain negative particles. He inferred there was a lump of positively charged material, with negative electrons throughout. He used the Raisins Bun Model to explain.
</span>Chadwick <span>proved that it consisted of a neutral particle with about the same mass as a proton "Neutron" is the name given to the particle</span>
Bohr believed Rutherford's prediction was correct, but it wasn't complete. Bohr proposed electrons could only move between energy levels, rather then being able to move everywhere.
J. J. Thomson is the corect awncer
Answer is: four iron atoms <span>are reacting with every three molecules of oxygen.
Balanced chemical reaction: 4Fe + 3O</span>₂ → 2Fe₂O₃.
From chemical reaction: n(Fe) : n(O₂) = 4 : 3.
Ratios of atoms of iron and molecules of oxygen is 4 : 3..
There is four atoms of iron and six atoms of oxygen on both side of reaction.
Answer:
a. 50KCal
b. 400KCal
c. Same as (a) above
Explanation:
Given
To raise the temperature of 1kg of liquid water at 1°C requires 1KCal
To raise the temperature of 1kg of ice or water vapour by 1°C requires 0.5KCal
To melt 1kg of ice at 0°C requires 80KCal
To evaporate 1kg of liquid water sitting at 100°C requires 540KCal
a. How much heat is required to raise the temperature of 5 kg of liquid water by 20 C?
To raise the temperature of 5 kg of ice by 20°C requires:
5 kg * (0.5 kcal / kgC) * 20C
= 50 KCal
b. How much heat is required to melt 5 kg of ice at 0 C?
To melt an ice of 5 kg of ice at 0 C requires:
5 kg * (80 kcal / kg)
= 400 KCal
c. Same as (a) above