An alkyne contains four carbon atoms.... so if you do 26 multiplied by 4 it equals 104... I do not know if that’s the answer so I apologize if it’s wrong :,)
        
             
        
        
        
The question incomplete , the complete question is:
A student dissolves of 18.0 g urea in 200.0 mL of a solvent with a density of 0.95 g/mL . The student notices that the volume of the solvent does not change when the urea dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to significant digits.
Answer:
The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.
Explanation:
Moles of urea = 
Volume of the solution = 200.0 mL = 0.2 L (1 mL = 0.001 L)

Molarity of the urea solution ;

Mass of solvent = m
Volume of solvent = V = 200.0 mL
Density of the urea = d = 0.95 g/mL


(1 g = 0.001 kg)
Molality of the urea solution ;


The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.
 
        
             
        
        
        
Explanation:
Reaction equation is as follows.
       
Here, 1 mole of  produces 2 moles of cations.
 produces 2 moles of cations.
![[Na^{+}] = 2[Na_{2}SO_{3}] = 2 \times 0.58](https://tex.z-dn.net/?f=%5BNa%5E%7B%2B%7D%5D%20%3D%202%5BNa_%7B2%7DSO_%7B3%7D%5D%20%3D%202%20%5Ctimes%200.58)
                                   = 1.16 M
![[SO^{2-}_{3}] = [Na_{2}SO_{3}]](https://tex.z-dn.net/?f=%5BSO%5E%7B2-%7D_%7B3%7D%5D%20%3D%20%5BNa_%7B2%7DSO_%7B3%7D%5D) = 0.58 M
 = 0.58 M
The sulphite anion will act as a base and react with  to form
 to form  and
 and  .
. 
As,     
                        = 
                        =  
 
According to the ICE table for the given reaction,
           
Initial:        0.58             0              0 
Change:     -x               +x             +x
Equilibrium: 0.58 - x     x               x
So, 
         ![K_{b} = \frac{[HSO^{-}_{3}][OH^{-}]}{[SO^{2-}_{3}]}](https://tex.z-dn.net/?f=K_%7Bb%7D%20%3D%20%5Cfrac%7B%5BHSO%5E%7B-%7D_%7B3%7D%5D%5BOH%5E%7B-%7D%5D%7D%7B%5BSO%5E%7B2-%7D_%7B3%7D%5D%7D)
  
         
                 x = 0.0003 M
So,   x = ![[HSO^{-}_{3}] = [OH^{-}]](https://tex.z-dn.net/?f=%5BHSO%5E%7B-%7D_%7B3%7D%5D%20%3D%20%5BOH%5E%7B-%7D%5D) = 0.0003 M
 = 0.0003 M
 ![[SO^{2-}_{3}]](https://tex.z-dn.net/?f=%5BSO%5E%7B2-%7D_%7B3%7D%5D) = 0.58 - 0.0003
 = 0.58 - 0.0003
                      = 0.579 M
Now, we will use ![[HSO^{-}_{3}]](https://tex.z-dn.net/?f=%5BHSO%5E%7B-%7D_%7B3%7D%5D) = 0.0003 M
 = 0.0003 M
The reaction will be as follows.
               
Initial:   0.0003
Equilibrium:  0.0003 - x        x             x
               
         
                       = 
                       = 
Therefore,  
As,  x <<<< 0.0003. So, we can neglect x.
Therefore,  
                               = 
                      x = 
 x = ![[OH^{-}] = [H_{2}SO_{3}]](https://tex.z-dn.net/?f=%5BOH%5E%7B-%7D%5D%20%3D%20%5BH_%7B2%7DSO_%7B3%7D%5D) =
 = 
     ![[H^{+}] = \frac{10^{-14}}{[OH^{-}]}](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%3D%20%5Cfrac%7B10%5E%7B-14%7D%7D%7B%5BOH%5E%7B-%7D%5D%7D)
                 = 
                 =  M
 M
Thus, we can conclude that the concentration of spectator ion is  M.
 M.
 
        
             
        
        
        
Answer:
True.
Explanation:
Here is an example: Hubble Space Telescope's launch in 1990 sped humanity to one of its greatest advances in that journey. Hubble is a telescope that orbits Earth. Its position above the atmosphere, which distorts and blocks the light that reaches our planet, gives it a view of the universe that typically far surpasses that of ground-based telescopes.
Hubble is one of NASA's most successful and long-lasting science missions. It has beamed hundreds of thousands of images back to Earth, shedding light on many of the great mysteries of astronomy. Its gaze has helped determine the age of the universe, the identity of quasars, and the existence of dark energy.