207 is the mass number. 82 would be the atomic number
Answer:
The correct answer is option E.
Explanation:
Structures for the reactants and products are given in an aimage ;
Number of double bonds in oxygen gas molecule = 1
Number of double bonds in nitro dioxide gas molecule = 1
Number of single bond in in nitro dioxide gas molecule = 1
Number of triple bonds in nitrogen gas molecule = 1
![\Delta H=[2 mol\times \Delta H_{f,NO_2}]-[1 mol\times \Delta H_{f,N_2}-2 mol\times \Delta H_{f,O_2}]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B2%20mol%5Ctimes%20%5CDelta%20H_%7Bf%2CNO_2%7D%5D-%5B1%20mol%5Ctimes%20%5CDelta%20H_%7Bf%2CN_2%7D-2%20mol%5Ctimes%20%5CDelta%20H_%7Bf%2CO_2%7D%5D)
(pure element)
(pure element )
The enthalpy of the given reaction is 15.86 kcal.
Answer:
<h2>1.89 atm</h2>
Explanation:
The new volume can be found by using the formula for Boyle's law which is
Since we are finding the new volume
From the question we have
We have the final answer as
<h3>1.89 atm</h3>
Hope this helps you
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:
Explanation:
Combustion. Have fun with that.
