Answer:
This unit has encouraged a deeper understanding of the world and it's guiding principles. While it was initially challenging for me to determine if a change was physical or chemical, this unit provided me with the information necessary to determine the type. With this knowledge, I can now interrelate with other properties and believe that this new ability will assist in future units as well. Thank you!
Explanation:
Answer:
a. A beta particle has a negative charge. d. A beta particle is a high-energy electron.
Explanation:
Identify the correct descriptions of beta particles.
a. A beta particle has a negative charge. YES. A beta particle is originated in the following nuclear reaction: ¹₀n ⇒ ¹₁H + ⁰₋₁e (beta particle.)
b. A beta particle contains neutrons. NO. It is a electron originated in the nucleus.
c. A beta particle is less massive than a gamma ray. NO. Gamma rays don't have mass while a beta particle has a mass which is half of one thousandth of the mass of a proton.
d. A beta particle is a high-energy electron. YES. Beta particles are nuclear originated hig-energy electrons.
Answer:
There will be one Al3+ ion.
There will be 3 NO3- ions
Explanation:
Dissociation equation:
Al(NO₃)₃ → Al³⁺ + 3NO₃¹⁻
When aluminium nitrate dissociate it produces one silver ion (Al³⁺) and three (NO₃¹⁻) ions.
Properties of Al(NO₃)₃:
It is inorganic compound having molecular mass 169.87 g/mol.
It is white odor less compound.
Its density is 4.35 g/mL.
Its melting and boiling points are 120°C and 440°C.
It is soluble in water.
It is sued to treat infections.
It is used in the photographic films.
It s toxic and must be handled with great care.
Answer:
1.30464 grams of glucose was present in 100.0 mL of final solution.
Explanation:
Moles of glucose = 
Volume of the solution = 100 mL = 0.1 L (1 mL = 0.001 L)
Molarity of the solution = 
A 30.0 mL sample of above glucose solution was diluted to 0.500 L:
Molarity of the solution before dilution = 
Volume of the solution taken = 
Molarity of the solution after dilution = 
Volume of the solution after dilution= 
Mass glucose are in 100.0 mL of the 0.07248 mol/L glucose solution:
Volume of solution = 100.0 mL = 0.1 L
Moles of glucose = 
Mass of 0.007248 moles of glucose :
0.007248 mol × 180 g/mol = 1.30464 grams
1.30464 grams of glucose was present in 100.0 mL of final solution.
