m moles of \(HNO_3=800×0.5\)
Moles of \(HNO_3=400×10^{−3}=0.4\) moles
Weight of \(HNO_3 = 0.4 × 63 g= 25.2\) g
Remaining acid \(= 25.2 – 11.5= 13.7\) g
\(M=\frac {13.7×1000}{400×63}\)
\(M=\frac {137}{252}\)
\(M=0.54\)
\(M=54×10^{−2}\)
Given that, The molarity of the remaining nitric acid solution is \(x×10^{−2} M\).
On comparing, \(x= 54\)
So, the answer is \(54\).
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32
Read More: Some Basic Concepts of Chemistry
There are two ways of classifying the matter:
Matter can exist in three physical states:
Based upon the composition, matter can be divided into two main types: