Answer:
The initial temperature is 499 K
Explanation:
Step 1: Data given
initial volume = 12 cm3 = 12 mL
Final volume = 7 cm3 = 7mL
The final temperature = 18 °C = 291 K
Step 2: Calculate the initial temperature
V1/T1 = V2/T2
⇒with V1 = the initial volume = 0.012 L
⇒with T1 = the initial volume = ?
⇒with V2 = the final volume 0.007 L
⇒with T2 = The final temperature = 291 K
0.012 / T1 = 0.007 / 291
0.012/T1 = 2.4055*10^-5
T1 = 0.012/2.4055*10^-5
T1 = 499 K
The initial temperature is 499 K
Answer:
29.75 Kg of NH₃
Solution:
In order to calculate the theoretical yield, first we will identify the limiting reactant.
According to equation,
6 g (3 moles) H₂ requires = 28 g (1 mole) N₂
So,
5250 g H₂ will require = X g of N₂
Solving for X,
X = (5250 g × 28 g) ÷ 6 g
X = 25433 g of N₂
Hence, it is found that H₂ is the limiting reactant because N₂ is provided in excess (32700 g). Therefore,
As,
6 g (3 mole) H₂ produced = 34 g ( 2 moles) of NH₃
So,
5250 g H₂ will produce = X g of NH₃
Solving for X,
X = (5250 g × 34 g) ÷ 6 g
X = 29750 g of NH₃
Or,
X = 29.75 Kg of NH₃
Answer:
a), b), c) & f)
Explanation:
d) does not apply because Ms value can be either +½ or -½
e) does not apply because Ml - values range from -l to +l, hence l= 2 doesn't exist when l= 1
they are male
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