Answer: 2800 calories
Explanation:
Latent heat of fusion is the amount of heat required to convert 1 mole of solid to liquid at atmospheric pressure.
Amount of heat required to fuse 1 gram of water = 80 cal
Mass of ice given = 35 gram
Heat required to fuse 1 g of ice at 
= 80 cal
Thus Heat required to fuse 35 g of ice =
Thus 2800 calories of energy is required to melt 35 g ice cube
Answer:
Why do most atoms form chemical bonds? They want a full outer shell of electrons, so the lose, gain, or share electrons with other elements, forming compounds, until they have 8 valence electrons and become stable. Double and triple covalent bonds that have greater bond energy and are shorter than single bonds.
Explanation: HOPE THIS HELPS YOU..
Answer:
Some of the physical changes used by the industrial chemist in order to identify it is by scratching it with other metals in order to find the hardness of it. Trying to deform it in order to find the malleability, and to heat it and measure the temperature in order to find the melting point.
Some of the chemical changes used by the industrial chemist in order to identify it is by inserting it in water to observe that whether it reacts with it or not, if the reaction is violent, then the metal belongs to either group I or group II. The other method is to insert it in acids of distinct strength and to observe its reaction. The metals belonging to the second group react briskly with acids. The other metals react gradually with acids and others are almost inert.
Remembering the periodic table, Fe is iron and O is oxygen. The elements that combine to create rust are iron and oxygen, with the chemical compound being (Iron)2(Oxygen)3.
Answer:
The average atomic weight = 121.7598 amu
Explanation:
The average atomic weight of natural occurring antimony can be calculated as follows :
To calculate the average atomic mass the percentage abundance must be converted to decimal.
121 Sb has a percentage abundance of 57.21%, the decimal format will be
57.21/100 = 0.5721 . The value is the fractional abundance of 121 Sb .
123 Sb has a percentage abundance of 42.79%, the decimal format will be
42.79/100 = 0.4279. The value is the fractional abundance of 123 Sb .
Next step is multiplying the fractional abundance to it masses
121 Sb = 0.5721 × 120.904 = 69.169178400
123 Sb = 0.4279 × 122.904 = 52.590621600
The final step is adding the value to get the average atomic weight.
69.169178400 + 52.590621600 = 121.7598 amu
