Concentrated nitric acid is labelled as 75% by mass. The volume in mL of the solution which contains 30 g of nitric acid is:
Given: Density of nitric acid solution is 1.25 g/mL.
We are given that the concentration of the nitric acid solution is 75% by mass. This means that for every 100 g of solution, there are 75 g of nitric acid. We are asked to find the volume of the solution that contains 30 g of nitric acid. First, calculate the total mass of the solution required to get 30 g of nitric acid: \[ {Mass of solution} = \frac{30 \, {g}}{0.75} = 40 \, {g} \] Now, using the density of the solution, which is 1.25 g/mL, calculate the volume of the solution: \[ {Volume} = \frac{40 \, {g}}{1.25 \, {g/mL}} = 32 \, {mL} \] Thus, the volume required is 40 mL.
Arrange the following in increasing order of solubility product:
\[ {Ca(OH)}_2, {AgBr}, {PbS}, {HgS} \]
Match List - I with List - II.
List - I (Saccharides) List - II (Glycosidic linkages found)
(A) Sucrose (I) \( \alpha 1 - 4 \)
(B) Maltose (II) \( \alpha 1 - 4 \) and \( \alpha 1 - 6 \)
(C) Lactose (III) \( \alpha 1 - \beta 2 \)
(D) Amylopectin (IV) \( \beta 1 - 4 \)
Choose the correct answer from the options given below:
Match List - I with List - II.
List - I (Complex) | List - II (Hybridisation) |
---|---|
(A) \([\text{CoF}_6]^{3-}\) | (I) \( d^2 sp^3 \) |
(B) \([\text{NiCl}_4]^{2-}\) | (II) \( sp^3 \) |
(C) \([\text{Co(NH}_3)_6]^{3+}\) | (III) \( sp^3 d^2 \) |
(D) \([\text{Ni(CN}_4]^{2-}\) | (IV) \( dsp^2 \) |
Choose the correct answer from the options given below:
Let \( A = \{-3, -2, -1, 0, 1, 2, 3\} \). A relation \( R \) is defined such that \( xRy \) if \( y = \max(x, 1) \). The number of elements required to make it reflexive is \( l \), the number of elements required to make it symmetric is \( m \), and the number of elements in the relation \( R \) is \( n \). Then the value of \( l + m + n \) is equal to:
For hydrogen-like species, which of the following graphs provides the most appropriate representation of \( E \) vs \( Z \) plot for a constant \( n \)?
[E : Energy of the stationary state, Z : atomic number, n = principal quantum number]
The number of 6-letter words, with or without meaning, that can be formed using the letters of the word MATHS such that any letter that appears in the word must appear at least twice, is $ 4 \_\_\_\_\_$.