What volume (in mL) of a 0.200 MHNO3 solution is required to completely react with 27.6 mL of a 0.100 MNa2CO3 solution according
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Answer:
Explanation:
To convert from moles to grams, we must use the molar mass.
Recall that water's molecular formula is H₂O. It contains hydrogen and oxygen. Look up the two elements masses on the Periodic Table.
- Hydrogen (H): 1.008 g/mol
- Oxygen (O): 15.999 g/mol
Now, use these masses to find water's mass. The subscript of 2 tells us there are 2 atoms of hydrogen, so we multiply hydrogen's mass by 2 and add oxygen's.
- H₂O= 2(1.008 g/mol) + 15.999 g/mol = 18.015 g/mol
Use the molar mass as a ratio.
Multiply by the given number of moles.
The moles of water will cancel.
Round to the nearest whole number. The 0 in the tenth place tells us to leave the number as is.
There are about <u>54 grams</u> of water in 3 moles.
Answer : The partial pressure of at equilibrium are, 1.133, 2.009, 0.574 bar respectively. The total pressure at equilibrium is, 3.716 bar
Solution : Given,
Initial pressure of = 1.42 bar
Initial pressure of = 2.87 bar
= 0.036
The given equilibrium reaction is,
Initially 1.42 2.87 0
At equilibrium (1.42-x) (2.87-3x) 2x
The expression of will be,
Now put all the values of partial pressure, we get
By solving the term x, we get
From the values of 'x' we conclude that, x = 3.889 can not more than initial partial pressures. So, the value of 'x' which is equal to 3.889 is not consider.
Thus, the partial pressure of at equilibrium = 2x = 2 × 0.287 = 0.574 bar
The partial pressure of at equilibrium = (1.42-x) = (1.42-0.287) = 1.133 bar
The partial pressure of at equilibrium = (2.87-3x) = [2.87-3(0.287)] = 2.009 bar
The total pressure at equilibrium = Partial pressure of + Partial pressure of
+ Partial pressure of
The total pressure at equilibrium = 1.133 + 2.009 + 0.574 = 3.716 bar