1. Given the following equation: N2 (g) + 3 H2 (g) ↔ 2NH3 (g) ΔH = -92 kJ/mol
a. this reaction is exothermic as ΔH is -ve
b. the equilibrium will shift 2 the left if nitrogen gas is removed
c. the equilibrium shift 2 the right if the temperature is lowered
d. the equilibrium shift 2 the left if ammonia (NH3) is added
e. principle of thermodynamic potential or Gibbs energy is used to answer B-D
Answer:
it may be electron, atom or ions depend on the nature of the substance and the character of reaction
like :- particles per mole, coulombs per electron
Answer:
Option A. 107 mL
Explanation:
From the question given above, the following data were obtained:
Initial volume (V₁) = 150 mL
Initial pressure (P₁) = 500 mmHg
Final pressure (P₂) = 700 mmHg
Temperature = constant
Final volume (V₂) =?
The final volume of the gas can be obtained by using the Boyle's law equation as shown below:
P₁V₁ = P₂V₂
500 × 150 = 700 × V₂
75000 = 700 × V₂
Divide both side by 700
V₂ = 75000 / 700
V₂ = 107 mL
Therefore, the final volume of the gas is 107 mL.
Answer:
Your strategy here will be to use the molar mass of potassium bromide,
KBr
, as a conversion factor to help you find the mass of three moles of this compound.
So, a compound's molar mass essentially tells you the mass of one mole of said compound. Now, let's assume that you only have a periodic table to work with here.
Potassium bromide is an ionic compound that is made up of potassium cations,
K
+
, and bromide anions,
Br
−
. Essentially, one formula unit of potassium bromide contains a potassium atom and a bromine atom.
Use the periodic table to find the molar masses of these two elements. You will find
For K:
M
M
=
39.0963 g mol
−
1
For Br:
M
M
=
79.904 g mol
−
1
To get the molar mass of one formula unit of potassium bromide, add the molar masses of the two elements
M
M KBr
=
39.0963 g mol
−
1
+
79.904 g mol
−
1
≈
119 g mol
−
So, if one mole of potassium bromide has a mas of
119 g
m it follows that three moles will have a mass of
3
moles KBr
⋅
molar mass of KBr
119 g
1
mole KBr
=
357 g
You should round this off to one sig fig, since that is how many sig figs you have for the number of moles of potassium bromide, but I'll leave it rounded to two sig figs
mass of 3 moles of KBr
=
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
360 g
a
a
∣
∣
−−−−−−−−−
Explanation:
<em>a</em><em>n</em><em>s</em><em>w</em><em>e</em><em>r</em><em>:</em><em> </em><em>3</em><em>6</em><em>0</em><em> </em><em>g</em><em> </em>
