Answer:
The formula for the anhydrous compound that was part of the mixture called natron that was used by the Egyptians is Na2(CO3)10(H2O).
They use this compound for medicine, cookery, agriculture, in glass-making and to dehydrate egyptian mummies.
Compound of sodium carbonate and sodium bicarbonate was the name of the resulting hydrate that formed.
Answer:
could be a that the apple produced a seed with genetic mutation in chromosome during the process meiosis.
Explanation:
Answer: -112200J
Explanation:
The amount of heat (Q) released from an heated substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)
Thus, Q = MCΦ
Since,
Q = ?
Mass of water vapour = 30.0g
C = 187 J/ G°C
Φ = (Final temperature - Initial temperature)
= 100°C - 120°C = -20°C
Then apply the formula, Q = MCΦ
Q = 30.0g x 187 J/ G°C x -20°C
Q = -112200J (The negative sign does indicates that heat was released to the surroundings)
Thus, -112200 joules of heat is released when cooling the superheated vapour.
Answer: Boron is the element which has properties of both metals and nonmetals.
Explanation:
Metals are defined as the elements which loose electrons to attain stable electronic configuration. They attain positive charge and form cation. Example: Zinc (Zn), Aluminium (Al) , copper (Cu)
Non-metals are defined as the elements which gain electrons to attain stable electronic configuration. They attain negative charge and form anion. Example: Chlorine (Cl) , Sulphur (S)
Metalloids are defined as the elements which show properties of both metals and non-metals. There are 7 metalloids in the periodic table. They are Boron (B) , Silicon (Si) , Germanium (Ge) , Arsenic (As) , Antimony (Sb), Tellurium (Te) and Polonium (Po).
Thus boron is the element which has properties of both metals and nonmetals.
Answer:

Explanation:
Data:
50/50 ethylene glycol (EG):water
V = 4.70 gal
ρ(EG) = 1.11 g/mL
ρ(water) = 0.988 g/mL
Calculations:
The formula for the boiling point elevation ΔTb is

i is the van’t Hoff factor — the number of moles of particles you get from 1 mol of solute. For EG, i = 1.
1. Moles of EG

2. Kilograms of water

3. Molal concentration of EG

4. Increase in boiling point

5. Boiling point