Yes, they are right.Very good hand writing! You did a very great job on showing your work.
Many of the atomic masses listed on the periodic table are decimal values because it's the weighted average mass of all of the isotopes of that element.
Answer:
Is water that is completely free (for the most part) of dissolved minerals.
Explanation:
It doesn't have most of the minerals, they can take it away by using a process called "Distillation." They boil the water to separate it from other chemical or mixtures in it. It is a process to clean the water.
By using ICE table:
CH3NH3+ + H2O → CH3NH4 2+ + OH-
initial 0.175 0 0
change -X +X +X
Equ (0.175-X) X X
when: Ka = Kw / Kb
= (1 x 10^-14) / (4.4 x 10^-4) = 2.3 x 10^-11
when Ka = [CH3NH42+][OH-] / [CH3NH3+]
by substitution:
2.3 x 10^-11 = X^2 / (0.175 - X ) by solving for X
∴ X = 2 x 10^-6
∴[OH-] = 2 x 10^-6
∴POH = -㏒[OH-]
= -㏒(2 x 10^-6)
= 5.7
when PH + POH = 14
∴PH = 14 - 5.7 = 8.3
Answer:
The correct answer is 89.6 L
Explanation:
We have the following chemical equation and the molar masses for the reaction:
3H₂(g) + N₂ --> 2 NH₃
6 g 28 g 34 g
That means that 3 moles of H₂ (6 g) reacts with 1 mol of N₂ (28 g) and gives 2 moles of NH₃ (34 g). In order to calculate how many liters of NH₃ result from the reaction of 12 grams of H₂ and 28 grams of N₂, we have to first figure out which reactant is the <em>limiting reactant</em>. According to the equation, if 6 grams of H₂ reacts with 28 g of N₂, and we have 12 grams:
6 g H₂------- 28 g N₂
12 g H₂-------- X = 12 g H₂ x 28 g N₂/6 g H₂ = 56 g N₂
We need 56 g of N₂ but we have 28 g of N₂, so <em>N₂ is the limiting reactant</em>. With the limiting reactant we can calculate the moles of product (NH₃) we will obtain:
We have 28 g N₂ -----> 28 g/14 g/mol = 2 moles N₂
1 mol N₂ ----------- 2 moles NH₃
2 mol N₂ --------- X = 2 mol N₂ x 2 moles NH₃/1 mol N₂ = 4 mol NH₃
Finally, we convert the moles of NH₃ to liters:
1 mol gas at STP = 22.4 L
Liters NH₃ obtained = 4 moles NH₃ x 22.4 L/1 mol = 89.6 L