% error =
![\frac{|experimental - theoretical|}{theoretical}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%7Cexperimental%20-%20theoretical%7C%7D%7Btheoretical%7D%20)
x 100%
Experimental: 2.85
Actual (theoretical): 2.70
% error =
![\frac{2.85-2.70}{2.70}](https://tex.z-dn.net/?f=%20%5Cfrac%7B2.85-2.70%7D%7B2.70%7D%20)
x 100% = .055555 x 100% = 5.56%
N2 + 3H2 ----> 2NH3
! mole of nitrogen reacts with 3 moles of hydrogen to give 2 moles of ammonia.
So by proportion number of moles NH3 obtained from 0.60 moles of N2 =
0.6 *2 = 1.2 moles Answer
Answer:The crayosphere- it is the portion of the Earth's surface where water is in solid form for at least one month of the year..-has been shrinking in response to climate warming.The important outcome,however,is the change and the response the human social system (infrastructure,food ,water,recreation) will be have to that change...
I hope It helps
Molality is defined as the number of moles of solute present in 1L of solution.
Molality (M) = No of moles of solute / 1 L of solution
Litre of solution = No of moles of solute / Molality
Molality = 0.220M
No of moles= 0.0170 mol
Volume in L = 0.0170/0.220
= 0.0772 L
Volume in mL = 77.27 mL
What is mole?
- A mole is a very important unit of measurement used by chemists.
- moles of something means there are 602,214,076,000,000,000,000,000,000,000,000 number of something.
- Chemists need to measure very small things such as atoms, molecules and other particles in moles.
- 602,214,076,000,000,000,000,000 is called Avogadro's number, abbreviated as 6.02 x
![10^{23}](https://tex.z-dn.net/?f=10%5E%7B23%7D)
Therefore, Volume in ml of 0.220 m hbr solution is required to produce 0.0170 moles is 77.27 mL.
To know more about Molality, visit:
brainly.com/question/24065939
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Answer:
pH = 11.37
Explanation:
Sodium cyanide will dissociate into sodium ion and cyanide ion. This cyanide ion will get hydrolyzed.The ICE table for hydrolysis of cyanide ion is:
![CN^{-}+ H_{2}O----------->HCN + OH^{-}](https://tex.z-dn.net/?f=CN%5E%7B-%7D%2B%20H_%7B2%7DO-----------%3EHCN%20%2B%20OH%5E%7B-%7D)
Initial 0.265 0 0
Change -x +x +x
Equilibrium 0.265-x x x
![K=\frac{[HCN][OH^{-}}{[CN^{-}]}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BHCN%5D%5BOH%5E%7B-%7D%7D%7B%5BCN%5E%7B-%7D%5D%7D)
This K is Kb of HCN
Kb = Kw / Ka
![Kb=\frac{10^{-14}}{4.9X10^{-10}}=2.04X10^{-5}](https://tex.z-dn.net/?f=Kb%3D%5Cfrac%7B10%5E%7B-14%7D%7D%7B4.9X10%5E%7B-10%7D%7D%3D2.04X10%5E%7B-5%7D)
Putting values
![2.04X10^{-5}=\frac{x^{2} }{(0.265-x)}](https://tex.z-dn.net/?f=2.04X10%5E%7B-5%7D%3D%5Cfrac%7Bx%5E%7B2%7D%20%7D%7B%280.265-x%29%7D)
x can be ignored in denominator as Kb is very low
![2.04X10^{-5}=\frac{x^{2} }{(0.265)}](https://tex.z-dn.net/?f=2.04X10%5E%7B-5%7D%3D%5Cfrac%7Bx%5E%7B2%7D%20%7D%7B%280.265%29%7D)
x= 2.33X10⁻³ M = [OH⁻]
pOH = -log[OH⁻]
pOH = -log(2.33X10⁻³ )
pOH = 2.63
pH = 14- pOH = 14-2.63 = 11.37