Nitrogen combine with hydrogen to produce ammonia
at a
ratio:

Assuming that the reaction has indeed proceeded to completion- with all nitrogen used up as the question has indicated.
of hydrogen gas would have been consumed while
of ammonia would have been produced. The final mixture would therefore contain
Apply the ideal gas law to find the total pressure inside the container and the respective partial pressure of hydrogen and ammonia:
The molar solubility is 7.4×
M and the solubility is 7.4×
g/L .
Calculation ,
The dissociation of silver bromide is given as ,
→
+ 
S
- S S
Ksp = [
] [
] = [S] [ S ] = 
S = √ Ksp = √ 5. 5×
= 7.4×
The solubility =7.4×
g/L
The molar solubility is the solubility of one mole of the substance.
Since , one mole of
is dissociates and form one mole of each
and
ion . So, solubility is equal to molar solubility but unit is different.
Molar solubility = 7.4×
mol/L = 7.4×
M
To learn more about molar solubility ,
brainly.com/question/16243859
#SPJ4
Answer:
See explanation and image attached
Explanation:
The reaction of 1-bromo-2-tert-butylcyclohexane with potassium tert-butoxide is an elimination reaction that occurs by E2 mechanism.
The E2 reaction proceeds faster when the hydrogens are in an antiperiplanar position at an angle of 180 degrees.
This is only attainable in the trans isomer of 1-bromo-2-tert-butylcyclohexane. Hence trans 1-bromo-2-tert-butylcyclohexane reacts faster with potassium tert-butoxide
Answer:
Option-C (27.36% Na, 1.20% H, 14.30% C, and 57.14% O)
Explanation:
<em>Percent Composition</em> is defined as the <u><em>%age by mass of each element present in a compound</em></u>. Therefore, it is a relative amount of each element present in a compound.
Calculating Percent Composition of NaHCO₃:
1: Calculating Molar Masses of all elements present in NaHCO₃:
a) Na = 22.99 g/mol
b) H = 1.01 g/mol
c) C = 12.01 g/mol
d) O₃ = 16.0 × 3 = 48 g/mol
2: Calculating Molecular Mass of NaHCO₃:
Na = 22.99 g/mol
H = 1.01 g/mol
C = 12.01 g/mol
O₃ = 48 g/mol
----------------------------------
Total 84.01 g/mol
3: Divide each element's molar mass by molar mass of NaHCO₃ and multiply it by 100:
For Na:
= 22.99 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100
= 27.36 %
For H:
= 1.01 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100
= 1.20 %
For C:
= 12.01 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100
= 14.29 % ≈ 14.30 %
For O:
= 48.0 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100
= 57.13 % ≈ 57.14 %
Stability of atoms is determined by neutron:proton ratio. This n/p ratio is 1:1 for elements below atomic number 20. Hope this helps.