The correct option is(A): 1.0 × 10–15
\(Bi_2S3⇋2Bi^3+2s+3s^2-3s\)
Ksp = (2s)2(3s)3 = 108s5
108s5 = 108 × 10–75
s = 1.0 × 10–15 mol/L




Let $ A \in \mathbb{R} $ be a matrix of order 3x3 such that $$ \det(A) = -4 \quad \text{and} \quad A + I = \left[ \begin{array}{ccc} 1 & 1 & 1 \\2 & 0 & 1 \\4 & 1 & 2 \end{array} \right] $$ where $ I $ is the identity matrix of order 3. If $ \det( (A + I) \cdot \text{adj}(A + I)) $ is $ 2^m $, then $ m $ is equal to:
A square loop of sides \( a = 1 \, {m} \) is held normally in front of a point charge \( q = 1 \, {C} \). The flux of the electric field through the shaded region is \( \frac{5}{p} \times \frac{1}{\varepsilon_0} \, {Nm}^2/{C} \), where the value of \( p \) is:
To neutralize stomach acids antacids type of drug are used. This tablet contain substances such as calcium, sodium bicarbonate, and magnesium, which act as base in the stomach.