<span>Avogadro's number
represents the number of units in one mole of any substance. This has the value
of 6.022 x 10^23 units / mole. This number can be used to convert the number of
atoms or molecules into number of moles. We calculate as follows:
2.30 g NH3 (1 mol / 17.03 g ) (</span>6.022 x 10^23 molecules / 1 mol ) = 8.13x10^22 molecules
The Correct option is E. Chemical energy to Mechanical energy.
Answer:
B) any complex thing with properties normally associated with living things
Answer:
0.4 M
Explanation:
Molarity is defined as moles of solute, which in your case is sodium hydroxide,
NaOH
, divided by liters of solution.
molarity
=
moles of solute
liters of solution
Notice that the problem provides you with the volume of the solution, but that the volume is expressed in milliliters,
mL
.
Moreover, you don't have the number of moles of sodium hydroxide, you just have the mass in grams. So, your strategy here will be to
determine how many moles of sodium hydroxide you have in that many grams
convert the volume of the solution from milliliters to liters
So, to get the number of moles of solute, use sodium hydroxide's molar mass, which tells you what the mass of one mole of sodium hydroxide is.
7
g
⋅
1 mole NaOH
40.0
g
=
0.175 moles NaOH
The volume of the solution in liters will be
500
mL
⋅
1 L
1000
mL
=
0.5 L
Therefore, the molarity of the solution will be
c
=
n
V
c
=
0.175 moles
0.5 L
=
0.35 M
Rounded to one sig fig, the answer will be
c
=
0.4 M
Explanation:
Answer:
D) He did not multiply the chlorine and oxygen atoms by the coefficient 4.
Explanation:
The coefficient 4 at the beginning of the chemical formula indicates that there are four Ca(ClO3)2 molecules. Think of this as Ca(ClO3)2 × 4. This means that he had to multiply the number of atoms for each element by 4 as well, so he should've ended up with 4 total calcium atoms (which is correct), 8 total chlorine atoms, and and 24 total oxygen atoms. He did not get all these answers because he didn't multiply the chlorine and oxygen atoms by the coefficient 4.
