Answer:
Mass = 0.697 g
Explanation:
Given data:
Volume of hydrogen = 1.36 L
Mass of ammonia produced = ?
Temperature = standard = 273.15 K
Pressure = standard = 1 atm
Solution:
Chemical equation:
3H₂ + N₂ → 2NH₃
First of all we will calculate the number of moles of hydrogen:
PV = nRT
R = general gas constant = 0.0821 atm.L/mol.K
1atm ×1.36 L = n × 0.0821 atm.L/mol.K × 273.15 K
1.36 atm.L = n × 22.43 atm.L/mol
n = 1.36 atm.L / 22.43 atm.L/mol
n = 0.061 mol
Now we will compare the moles of hydrogen and ammonia:
H₂ : NH₃
3 : 2
0.061 : 2/3×0.061 = 0.041
Mass of ammonia:
Mass = number of moles × molar mass
Mass = 0.041 mol × 17 g/mol
Mass = 0.697 g
Answer:
Hot and something you do not touch
Explanation:
Answer:
Buffer 1.
Explanation:
Ammonia is a weak base. It acts like a Bronsted-Lowry Base when it reacts with hydrogen ions.
.
gains one hydrogen ion to produce the ammonium ion
. In other words,
is the conjugate acid of the weak base
.
Both buffer 1 and 2 include
- the weak base ammonia
, and - the conjugate acid of the weak base
.
The ammonia
in the solution will react with hydrogen ions as they are added to the solution:
.
There are more
in the buffer 1 than in buffer 2. It will take more strong acid to react with the majority of
in the solution. Conversely, the pH of buffer 1 will be more steady than that in buffer 2 when the same amount of acid has been added.
Answer:
c = 0.13 j/ g.°C
Explanation:
Given data:
Mass of mercury = 29.5 g
Initial temperature = 32°C
Final temperature = 161°C
Heat absorbed = 499.2 j
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
Q = m.c. ΔT
ΔT = T2 - T1
ΔT = 161°C - 32°C
ΔT = 129 °C
Q = m.c. ΔT
c = Q / m. ΔT
c = 499.2 j / 29.5 g. 129 °C
c = 499.2 j / 3805.5 g. °C
c = 0.13 j/ g.°C
Answer:
2.14 moles of H₂O₂ are required
Explanation:
Given data:
Number of moles of H₂O₂ required = ?
Number of moles of N₂H₄ available = 1.07 mol
Solution:
Chemical equation:
N₂H₄ + 2H₂O₂ → N₂ + 4H₂O
now we will compare the moles of H₂O₂ and N₂H₄
N₂H₄ : H₂O₂
1 : 2
1.07 : 2×1.07 = 2.14 mol