Answer: The concentration of hydroxide ions at this temperature is 
Explanation:
When an expression is formed by taking the product of concentration of ions raised to the power of their stoichiometric coefficients in the solution of a salt is known as ionic product.
The ionic product for water is written as:
![K_w=[H^+][OH^-]](https://tex.z-dn.net/?f=K_w%3D%5BH%5E%2B%5D%5BOH%5E-%5D)
![5.13\times 10^{-13}=[H^+][OH^-]](https://tex.z-dn.net/?f=5.13%5Ctimes%2010%5E%7B-13%7D%3D%5BH%5E%2B%5D%5BOH%5E-%5D)
As ![[H^+]=[OH^-]](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D%5BOH%5E-%5D)


The concentration of hydroxide ions at this temperature is 
Water is an amphoteric compound. This means it could be a base or an acid, depending on the substance it is to be reacted with. In this case, water is a base because HF is an acid. Now, if the reactant is an acid, its form after the reaction is called the conjugate pair. Since HF became F⁻, <em>the acid-conjugate base pair is: HF and F⁻.</em>
Always younger than the rock it intrudes
Answer : Option a) True.
Explanation : A placebo is a substance that is unlikely to affect the dependent variable. True as in a clinical trial for any change in the placebo arm is known as the placebo response, and the difference between this placebo response and the result of no treatment is called as the placebo effect. Therefore, a placebo may be given to any person in a clinical context in order to deceive that recipient into thinking that it is an part of active treatment. It actually resembles a drug but is not an actual active drug, so, it is clear that it will not affect dependent variable.
Answer : The heat of combustion of n-propanol is 0.554 kJ/mol
Explanation :
First we have to calculate the moles of n-propanol.

Molar mass of n-propanol = 60.09 g/mole

Now we have to calculate the heat of combustion of n-propanol.
As, 0.0166 mole of n-propanol liberated heat of combustion = -33.4 kJ
So, 1 mole of n-propanol liberated heat of combustion = 0.0166 × (-33.4 kJ)
= 0.554 kJ/mol
Therefore, the heat of combustion of n-propanol is 0.554 kJ/mol