Answer:
89.6L
Explanation:
1mole of any gas occupies 22.4L. This simply means that,
1mole of CO2 occupies 22.4L at stp.
Therefore, 4moles of CO2 will occupy = 4 x 22.4 = 89.6L
PH= −log
10
[H
+
]
= −log
10
(0.001)
= −log
10
(10
−3
)
= −(−3)log
10
10
pH=3.
01
The anode is the electrode where the oxidation occurs.
Cathode is the electrode where the reducction occurs.
Equations:
Mn(2+) + 2e- ---> Mn(s) Eo = - 1.18 V
2Fe(3+) + 2e- ----> 2 Fe(2+) 2Eo = + 1.54 V
The electrons flow from the electrode with the lower Eo to the electrode with the higher Eo yielding to a positive voltage.
Eo = 1.54 V - (- 1.18) = 1.54 + 1.18 = 2.72
Answer: 2.72 V
Answer:
Carbohydrates can be divided into two main types: simple and complex. Simple carbohydrates are made up of just one or two sugar units, whereas complex carbohydrates are made up of many sugar units.
Answer:
2.57 g of H₂
Solution:
The Balance Chemical Equation is as follow,
N₂ + 3 H₂ → 2 NH₃
According to Balance equation,
34.06 g (2 moles) NH₃ is produced by = 6.04 g (3 moles) of H₂
So,
14.51 g of NH₃ will be produced by = X g of H₂
Solving for X,
X = (14.51 g × 6.04 g) ÷ 34.06 g
X = 2.57 g of H₂
