Methanol is prepared by reacting Carbon monoxide and Hydrogen gas,
CO + 2 H₂ → CH₃OH
Calculating Moles of CO:
According to equation,
32 g (1 mole) of CH₃OH is produced by = 1 Mole of CO
So,
3.60 × 10² g of CH₃OH is produced by = X Moles of CO
Solving for X,
X = (3.60 × 10² g × 1 Mole) ÷ 32 g
X = 11.25 Moles of CO
Calculating Moles of H₂:
According to equation,
32 g (1 mole) of CH₃OH is produced by = 2 Mole of H₂
So,
3.60 × 10² g of CH₃OH is produced by = X Moles of H₂
Solving for X,
X = (3.60 × 10² g × 2 Mole) ÷ 32 g
X = 22.5 Moles of H₂
Result:
3.60 × 10² g of CH₃OH is produced by reacting 11.25 Moles of CO and 22.5 Moles of H₂.
Answer:
Hello friends
Explanation:
<h3>For a given principal quantum number for or n, the corresponding angular quantum number or is equivalent to a range between 0 and( n-1)</h3>
<h3>This means that the angular quantum number for a principal quantum number of 2 is equivalent to.</h3>
<h3>1 = 0 - > (n - 1) = 0 - > (2 - 1) = 0 - > 1</h3>
<h3>Hope it's helpfully. </h3>
Answer:
The final volume is 1.6 L.
Explanation:
It is given that,
A diver has a lung capacity of 2.4 L when the pressure is 0.8 atm. We need to find the volume of the diver’s lungs when the pressure changes to 1.2 atm. Let V₂ is volume.
It is based on Boyle's law. According to this law,

K is constant


So, the final volume is 1.6 L.
Explanation:
(a) mass number = atomic number + number of neutrons
(the atomic number is the number of protons of an element)
mass number of A = 11 + 12 = 23
mass number of B = 17 + 18 = 35
(b) +1 and -1, respectively
(c) A+ + B- → AB (the first plus sign and the minus sign are superscripts)