The addition of hydrogen carbonate to bromothymol blue turns the solution blue. Thus, option B is correct.
The balanced equation for the dissociation of bromothymol blue is:
The color of dissociated form is yellow and undissociated form is blue.
<h3>What is the final color of solution?</h3>
The addition of hydrogen carbonate results in the dissociated ions as:
The dissociation results in the increased hydrogen ion concentration. The undissociated form in the reaction mixture increases.
Thus, the color of the solution will turn blue. Hence, option B is correct.
Learn more about bromothymol blue, here:
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Answer:
The value of Kc is 660 (Option E) is correct
Explanation:
Step 1: Data given
Kp = 27
Temperature = 25.0 °C
Step 2: The balanced equation
2 NO(g) + Br2(g) ⇄ 2 NOBr(g)
Step 3: Calculate Kc
Kp = Kc * (RT)^Δn
⇒ with Kp = 27
⇒ with Kc = TO BE DETERMINED
⇒ with R = the gas constant = 0.08206 L*atm/mol*K
⇒ with T = The temperature = 25 °C = 298 K
⇒ Δn = the difference in moles = -1
27 = Kc * (0.08206*298)^-1
Kc = 660
The value of Kc is 660 (Option E) is correct
Answer:
The equilibrium concentration of CH₃OH is 0.28 M
Explanation:
For the reaction: CO (g) + 2H₂(g) ↔ CH₃OH(g)
The equilibrium constant (Keq) is given for the following expresion:
Keq= 
=14.5
Where (CH3OH), (CO) and (H2) are the molar concentrations of each product or reactant.
We have:
(CH3OH)= ?
(CO)= 0.15 M
(H2)= 0.36 M
So, we only have to replace the concentrations in the equilibrium constant expression to obtain the missing concentration we need:
14.5= 
14.5 x (0.15 M) x 
= (CH₃OH)
0.2818 M = (CH₃OH)
<span>Carbon can also bond with other
four atoms because of its outer shell (valence shell) that has four electrons.
This is the reason why organic molecules can be so large because of this
bonding. Suppose you have a compound of CCl4. You know that chlorine can only
share 1 electron because 7 of its electrons are filled. Also, in carbon, it can
only share 4 electrons because 4 of it are already filled. That is why carbon
needs four chlorine to form CCl4. The answer is letter <u>B.</u></span>
