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 \(54\).
If the system of equations \[ x + 2y - 3z = 2, \quad 2x + \lambda y + 5z = 5, \quad 14x + 3y + \mu z = 33 \] has infinitely many solutions, then \( \lambda + \mu \) is equal to:}
The equilibrium constant for decomposition of $ H_2O $ (g) $ H_2O(g) \rightleftharpoons H_2(g) + \frac{1}{2} O_2(g) \quad (\Delta G^\circ = 92.34 \, \text{kJ mol}^{-1}) $ is $ 8.0 \times 10^{-3} $ at 2300 K and total pressure at equilibrium is 1 bar. Under this condition, the degree of dissociation ($ \alpha $) of water is _____ $\times 10^{-2}$ (nearest integer value). [Assume $ \alpha $ is negligible with respect to 1]
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: