Answer:
7 gram of Na2SO4 should be required to prepare 0.5L of a 0.100 M solution.
Explanation:
First of all the molecular weight of Na2SO4 is 142.08 gram.Now we all know that if the molecular weight of a compound is dissolved in 1000ml or 1 litee of water then the strength of that solution becomes 1 M.
According to the given question we have to prepare 0.100 M solution
1000 ml of solution contain 142.08×0.1= 14.208 gram Na2SO4
1 ml of solution contain 14.208÷1000= 0.014 gram
0.5L or 500ml of solution contain 0.014×500= 7gram Na2SO4.
So it can be stated that 7 gram of Na2SO4 should be required to prepare 0.5L of a 0.100M solution.
Answer:The equilibrium constant for a given reaction is [concentration of products]/[concentration of reactants].
Explanation:
Equilibrium constant=[concentration of products]/[concentration of reactants]
The concentration of reactant molecules is maximum at time 0 and it decreases as the reaction proceeds, The concentration of product molecules increases.At equilibrium the concentration of reactants and products are equal.
All the changes would occur in accordance with the LeChateliers principle.
For the given reaction the following changes would occur:
a When CO is removed from the reaction mixture so the reaction would shift towards right that is in forward direction as we are decreasing the concentration of CO so the system would try to increase the concentration of CO and that can happen by more production of CO.
b Since the above reaction is an endothermic reaction so when we would be adding heat to the system that is when we would increase the temperature the reaction would shift forwards as more heat energy is absorbed by reactants to form more products.
c When more CO₂ is added so more amount of reactants are added to the system so the system would try to decrease the amount of reactants that is CO₂ and hence more amount of products would be formed.The reaction would shift in forward direction.
d Since this reaction is endothermic in nature so when we remove the heat from reaction hence even less amount of heat is present in the system and so the reaction shift in backward direction as the reaction cannot proceed without enough amount of heat.
Surface tension under water results from greater attraction of liquid molecules to each other, due to a process called cohesion, than to molecules in the air, due to a process called adhesion.
Answer:
Concentration of OH⁻:
1.0 × 10⁻⁹ M.
Explanation:
The following equilibrium goes on in aqueous solutions:
.
The equilibrium constant for this reaction is called the self-ionization constant of water:
![K_w = [\text{H}^{+}]\cdot[\text{OH}^{-}]](https://tex.z-dn.net/?f=K_w%20%3D%20%5B%5Ctext%7BH%7D%5E%7B%2B%7D%5D%5Ccdot%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D)
.
Note that water isn't part of this constant.
The value of 
at 25 °C is 
. How to memorize this value?
- The pH of pure water at 25 °C is 7.
![[\text{H}^{+}] = 10^{-\text{pH}} = 10^{-7}\;\text{mol}\cdot\text{dm}^{-3}](https://tex.z-dn.net/?f=%5B%5Ctext%7BH%7D%5E%7B%2B%7D%5D%20%3D%2010%5E%7B-%5Ctext%7BpH%7D%7D%20%3D%2010%5E%7B-7%7D%5C%3B%5Ctext%7Bmol%7D%5Ccdot%5Ctext%7Bdm%7D%5E%7B-3%7D)
- However,
![[\text{OH}^{-}] = [\text{H}^{+}]=10^{-7}\;\text{mol}\cdot\text{dm}^{-3}](https://tex.z-dn.net/?f=%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D%20%3D%20%5B%5Ctext%7BH%7D%5E%7B%2B%7D%5D%3D10%5E%7B-7%7D%5C%3B%5Ctext%7Bmol%7D%5Ccdot%5Ctext%7Bdm%7D%5E%7B-3%7D)
for pure water. - As a result,
![K_w = [\text{H}^{+}] \cdot[\text{OH}^{-}] = (10^{-7})^{2} = 10^{-14}](https://tex.z-dn.net/?f=K_w%20%3D%20%5B%5Ctext%7BH%7D%5E%7B%2B%7D%5D%20%5Ccdot%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D%20%3D%20%2810%5E%7B-7%7D%29%5E%7B2%7D%20%3D%2010%5E%7B-14%7D)
at 25 °C.
Back to this question. ![[\text{H}^{+}]](https://tex.z-dn.net/?f=%5B%5Ctext%7BH%7D%5E%7B%2B%7D%5D)
is given. 25 °C implies that 
. As a result,
![\displaystyle [\text{OH}^{-}] = \frac{K_w}{[\text{H}^{+}]} = \frac{10^{-14}}{1.0\times 10^{-5}} = 10^{-9} \;\text{mol}\cdot\text{dm}^{-3}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D%20%3D%20%5Cfrac%7BK_w%7D%7B%5B%5Ctext%7BH%7D%5E%7B%2B%7D%5D%7D%20%3D%20%5Cfrac%7B10%5E%7B-14%7D%7D%7B1.0%5Ctimes%2010%5E%7B-5%7D%7D%20%3D%2010%5E%7B-9%7D%20%5C%3B%5Ctext%7Bmol%7D%5Ccdot%5Ctext%7Bdm%7D%5E%7B-3%7D)
.
