answer:To find the number of moles of NaHSO4 formed when 2 moles of NaOH react with 2 moles of H2SO4, we need to look at the balanced chemical equation for the reaction: NaOH + H2SO4 → NaHSO4 + H2O From the balanced equation, we can see that 1 mole of NaOH reacts with 1 mole of H2SO4 to produce 1 mole of NaHSO4 and 1 mole of H2O. Given that we have 2 moles of NaOH and 2 moles of H2SO4, the reaction would proceed as follows: 2 NaOH + 2 H2SO4 → 2 NaHSO4 + 2 H2O Since the stoichiometry of the reaction is 1:1 for NaOH and H2SO4 to produce NaHSO4, and we have equal moles of NaOH and H2SO4 (2 moles each), they will react completely to form 2 moles of NaHSO4. Therefore, the number of moles of NaHSO4 formed is boxed{2} moles.

answer:1. **Restate the Problem:** We have 2008 white stones and 1 black stone in a row. An 'action' means selecting one black stone and changing the color of neighboring stone(s). We need to find all possible initial positions of the black stone to make all stones black by finite actions. 2. **Generalize the Problem:** Suppose instead of 2009 stones, we have ( n in 2mathbb{N}+1 ) stones. Due to symmetry, we only need to consider one side of the row. Number the stones from 1 to ( n ). 3. **Algorithm Description:** The algorithm looks at the first black stone and performs the action described. It tries to fill in the white stones closest to an end, making them black. If successful, it moves to the other end and fills any gaps in the consecutive array of black stones. 4. **Case Analysis:** - **Case 1:** ( i equiv 0 pmod{2} ) - Example configuration: ( boxed{0} boxed{0} boxed{0} boxed{1} boxed{0} boxed{0} boxed{0} boxed{0} boxed{0} boxed{0} boxed{0} boxed{0} boxed{0} ) - After some moves, we get: ( boxed{1} boxed{1} boxed{1} boxed{1} boxed{1} boxed{1} boxed{1} boxed{0} boxed{0} boxed{0} boxed{0} boxed{0} boxed{0} ) - The algorithm continues to push forward, but eventually fails as it cannot recover from having the first stone white. - **Case 2:** ( i equiv 1 pmod{2} ) - Similar logic applies, and the algorithm terminates faster. 5. **Central Stone Analysis:** - When ( i = frac{n+1}{2} ), we use induction to prove it works. - **Base Case:** ( n = 3 ) - Configuration: ( boxed{0} boxed{1} boxed{0} ) - One move makes all stones black. - **Inductive Step:** - Assume it works for ( n = k ). - Add one stone to each end: ( boxed{0} boxed{1} boxed{1} dots boxed{1} boxed{1} boxed{1} dots boxed{1} boxed{1} boxed{0} ) - Apply the algorithm to fill the first half, resulting in: ( boxed{0} boxed{1} boxed{0} boxed{1} boxed{1} dots boxed{1} boxed{1} ) - One move finishes the process. - By induction, if the central piece is black, the array can be filled. 6. **Conclusion:** - For ( n = 2009 ), the central piece is the 1005th stone.

answer:(1) Given that the asymptote lines of the hyperbola C are y=±x, and one of its semi-axes is x= frac{ sqrt{2}}{2}. We can deduce that a=b, frac{a^{2}}{c}= frac{ sqrt{2}}{2}, and frac{c}{a}= sqrt{2}. Solving these equations, we get a=b=1 and c= sqrt{2}. The equation of the hyperbola is: boxed{x^{2}-y^{2}=1}. (2) Let the equation of line l be: x=my-2. From the problem, we can obtain the following system of equations: begin{cases}x=my-2 x^{2}-y^{2}=1end{cases} Solving this system, we get (m^{2}-1)y^{2}-4my+3=0. Thus, y_{1}+y_{2}= frac{4m}{m^{2}-1} and y_{1}y_{2}= frac{3}{m^{2}-1} Also, |y_{1}-y_{2}|= sqrt{ frac{16m^{2}-12m^{2}+12}{(m^{2}-1)^{2}}}= frac{ sqrt{12+4m^{2}}}{|m^{2}-1|} The area of triangle OAB is 2 sqrt{3}. Therefore, frac{1}{2} times 2 times frac{ sqrt{12+4m^{2}}}{|m^{2}-1|}=2 sqrt{3} Solving this equation, we get m=± frac{ sqrt{21}}{3}. The equation of line l is: boxed{x=± frac{ sqrt{21}}{3}y-2}. (3) From part (2), we know that y_{1}y_{2}= frac{3}{m^{2}-1}. Also, x_{1}x_{2}=(my_{1}-2)(my_{2}-2)=m^{2}y_{1}y_{2}-2m(y_{1}+y_{2})+4=frac{3m^{2}}{m^{2}-1}-frac{8m^{2}}{m^{2}-1}+4=frac{-m^{2}-4}{m^{2}-1}. If OA perp OB, then frac{frac{3}{m^{2}-1}}{frac{-m^{2}-4}{m^{2}-1}}=-1. Solving this equation, we get m^{2}=-1. Since the square of a real number cannot be negative, such a line l does not exist.

answer:1. **Define Points and Intersections:** Let the lines (A A_1) and (C C_1) intersect at point (O). We need to prove that line (B B_1) passes through point (O) if and only if ( frac{CB_1}{B_1A} = frac{1}{pq} ). 2. **Assign Masses:** Place masses of (1), (p), and (pq) at points (A), (B), and (C) respectively. 3. **Find Centers of Mass:** - The center of mass of points (A) and (B) when (A) has mass (1) and (B) has mass (p) is point (C_1). - The center of mass of points (B) and (C) when (B) has mass (p) and (C) has mass (pq) is point (A_1). Therefore, the point (O), which is the intersection of lines (A A_1) and (C C_1), is the center of mass of points (A), (B), and (C) with the given masses. 4. **Verify the Intersection:** By the centers of mass, point (O) lies on the line connecting point (B) with the center of mass of points (A) and (C). 5. **Determine (B_1):** If (B_1) is the center of mass of points (A) and (C) with masses (1) and (pq), respectively, it satisfies: [ frac{AB_1}{B_1C} = pq : 1 ] Since there exists a unique point on segment (AC) that divides it in the ratio of (pq : 1), we conclude the location of (B_1). 6. **Conclusion:** Hence, the line (B B_1) passing through (O) implies that ( frac{CB_1}{B_1A} = frac{1}{pq} ). [ boxed{frac{CB_1}{B_1A} = frac{1}{pq}} ]