Answer:
100ml of a stock 50% KNO3 solutions are needed to prepare 250ml of a 20% KNO3 solution.
Explanation:
In the given question it is mentioned that
S1=50%
V2=250ml
S2= 20%
We all know that
V1S1=V2S2
∴V1= V2×S2÷S1
∴V1= V2S2×1/S1
∴V1= 250×20÷50
∴V1= 100ml
In a combustion reaction, one of the reactants is always oxygen.
Answer: The concentration of hydrogen ions for this solution is 
.
Explanation:
Given: pOH = 11.30
The relation between pH and pOH is as follows.
pH + pOH = 14
pH + 11.30 = 14
pH = 14 - 11.30
= 2.7
Also, pH is the negative logarithm of concentration of hydrogen ions.
![pH = - log [H^{+}]](https://tex.z-dn.net/?f=pH%20%3D%20-%20log%20%5BH%5E%7B%2B%7D%5D)
Substitute the values into above formula as follows.
![pH = -log [H^{+}]\\2.7 = -log [H^{+}]\\conc. of H^{+} = 1.99 \times 10^{-3}](https://tex.z-dn.net/?f=pH%20%3D%20-log%20%5BH%5E%7B%2B%7D%5D%5C%5C2.7%20%3D%20-log%20%5BH%5E%7B%2B%7D%5D%5C%5Cconc.%20of%20H%5E%7B%2B%7D%20%3D%201.99%20%5Ctimes%2010%5E%7B-3%7D)
Thus, we can conclude that the concentration of hydrogen ions for this solution is .
Answer:
(C) Acetylene (ethyne) can be converted to the acetylide anion by treating with a strong base such as CH₃Li.
Explanation:
Acetylene (C₂H₂) can be converted to the acetylide anion (C₂⁻²) when treated with a base because it will donate protons (2H⁺). So it will be a neutralization reaction. NaNH₂ and NaOH are strong bases because they are good electrons donators ( NaNH₂ has pair of electrons on N, and NaOH has the group OH⁻), but CH₃Li has no pair of electrons to donate, so it's not a strong base.
Answer:
Hey hi
Explanation:
Can you pls tell me which language is this pls I really want to help you... Sorry
