This is an incomplete question, here is a complete question.
What are the concentrations of hydrogen ion and hydroxide ion in household ammonia, an aqueous solution of NH₃ that has a pH of 11.00?
Answer : The concentrations of hydrogen ion and hydroxide ion in household ammonia is,
and ![1.00\times 10^{-3}M](https://tex.z-dn.net/?f=1.00%5Ctimes%2010%5E%7B-3%7DM)
Explanation :
As we know that the aqueous solution of ammonia is ammonium hydroxide that is, ![NH_4OH](https://tex.z-dn.net/?f=NH_4OH)
The equilibrium reaction will be:
![NH_4OH\rightleftharpoons NH_4^++OH^-](https://tex.z-dn.net/?f=NH_4OH%5Crightleftharpoons%20NH_4%5E%2B%2BOH%5E-)
Given:
pH = 11.00
First we have to calculate the hydrogen ion concentration.
pH : It is defined as the negative logarithm of hydrogen ion concentration.
![pH=-\log [H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH%5E%2B%5D)
![11.00=-\log [H^+]](https://tex.z-dn.net/?f=11.00%3D-%5Clog%20%5BH%5E%2B%5D)
![[H^+]=1.00\times 10^{-11}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D1.00%5Ctimes%2010%5E%7B-11%7DM)
Now we have to calculate the pOH.
![pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-11.00\\\\pOH=3.00](https://tex.z-dn.net/?f=pH%2BpOH%3D14%5C%5C%5C%5CpOH%3D14-pH%5C%5C%5C%5CpOH%3D14-11.00%5C%5C%5C%5CpOH%3D3.00)
Now we have to calculate the hydroxide ion concentration.
![pOH=-\log [OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%20%5BOH%5E-%5D)
![3.00=-\log [OH^-]](https://tex.z-dn.net/?f=3.00%3D-%5Clog%20%5BOH%5E-%5D)
![[OH^-]=1.00\times 10^{-3}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1.00%5Ctimes%2010%5E%7B-3%7DM)
Therefore, the concentrations of hydrogen ion and hydroxide ion in household ammonia is,
and ![1.00\times 10^{-3}M](https://tex.z-dn.net/?f=1.00%5Ctimes%2010%5E%7B-3%7DM)
If you look at your Chemistry reference table, the answer should be about 62g of NH4Cl. Lining up the lines of solubility and temperature, you should come to that answer.
Answer: a nucleic acid that contains the genetic instructions for the development and function of living things
Explanation:
Answer TRUE
Newton's first law of motion states that "An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force."
Answer:
The length of the cubical container will be 17.206 m.
Explanation:
Amount of carbon dioxide gas generated in year by person = 10,000 kg
1 kg = 1000 g
10,000 kg = 10,000 × 1000 g = 10,000,000 g
Moles of carbon dioxide = ![\frac{10,000,000 g}{44 g/mol}=227,272.72 mol](https://tex.z-dn.net/?f=%5Cfrac%7B10%2C000%2C000%20g%7D%7B44%20g%2Fmol%7D%3D227%2C272.72%20mol)
Volume of carbon dioxide gas at STP= V
STP conditions = P = 1 atm , T = 273 K
Using Ideal gas equation ;
![PV=nRT](https://tex.z-dn.net/?f=PV%3DnRT)
![V=\frac{nRT}{P}=\frac{227,272.72mol\times 0.0821 atm L/mol K\times 273 K}{1 atm}](https://tex.z-dn.net/?f=V%3D%5Cfrac%7BnRT%7D%7BP%7D%3D%5Cfrac%7B227%2C272.72mol%5Ctimes%200.0821%20atm%20L%2Fmol%20K%5Ctimes%20273%20K%7D%7B1%20atm%7D)
![V=5,093,931.82 L](https://tex.z-dn.net/?f=V%3D5%2C093%2C931.82%20L)
![1 L = 0.001 m^3](https://tex.z-dn.net/?f=1%20L%20%3D%200.001%20m%5E3)
![V=5,093,931.82 L=5,093,931.82 \times 0.001 m^3=5,093.93182 m^3](https://tex.z-dn.net/?f=V%3D5%2C093%2C931.82%20L%3D5%2C093%2C931.82%20%5Ctimes%200.001%20m%5E3%3D5%2C093.93182%20m%5E3)
Let the side of the cube = a
Volume of the cube = ![a^3](https://tex.z-dn.net/?f=a%5E3)
![V=5,093.93182 m^3](https://tex.z-dn.net/?f=V%3D5%2C093.93182%20m%5E3)
![a^3=5,093.93182 m^3](https://tex.z-dn.net/?f=a%5E3%3D5%2C093.93182%20m%5E3)
![a=\sqrt{5,093.93182 m^3}=17.206 m](https://tex.z-dn.net/?f=a%3D%5Csqrt%7B5%2C093.93182%20m%5E3%7D%3D17.206%20m)
The length of the cubical container will be 17.206 m.