Answer:
8,5 E10 nm
Explanation:
⇒ 0.85 E-2 Km * ( 1000 m / Km ) * ( 1 E9 nm / m ) = 8.5 E10 nm
⇒ 0.85 E-2 Km = 8.5 E10 nm
The balanced net equation for
BaCl2 (aq) + H2SO4(aq) → BaSO4(s) + HCl (aq) is
Ba^2+(aq) +SO4^2- → BaSO4 (s)
<u><em>Explanation</em></u>
Ionic equation is a chemical equation in which electrolytes in aqueous solution are written as dissociated ions.
<u>ionic equation is written using the below steps</u>
Step 1: <em>write a balanced molecular equation</em>
BaCl2 (aq) +H2SO4 (aq)→ BaSO4(s) +2HCl (aq)
Step 2: <em>Break all soluble electrolytes in to ions</em>
= Ba^2+ (aq) + 2Cl^-(aq) + 2H^+(aq) + SO4^2-(aq)→ BaSO4(s) + 2H^+(aq) +2Cl^- (aq)
step 3: <em>cancel the spectator ions in both side of equation ( ions which do not take place in the reaction)</em>
<em> </em><em> =</em> 2Cl^- and 2H^+ ions
Step 4: <em>write the final net equation</em>
<em> Ba^2+(aq) + SO4^2-(aq)→ BaSO4(s</em><em>)</em>
Answer:
Use a ratio of 0.44 mol lactate to 1 mol of lactic acid
Explanation:
John could prepare a lactate buffer.
He can use the Henderson-Hasselbalch equation to find the acid/base ratio for the buffer.
![\text{pH} = \text{pK}_{\text{a}} + \log\dfrac{\text{[A$^{-}$]}}{\text{[HA]}}\\\\3.5 = 3.86 + \log\dfrac{\text{[A$^{-}$]}}{\text{[HA]}}\\\\\log\dfrac{\text{[A$^{-}$]}}{\text{[HA]}} = 3.5 - 3.86 = -0.36\\\\\dfrac{\text{[A$^{-}$]}}{\text{[HA]}} = 10^{-0.36} = \mathbf{0.44}](https://tex.z-dn.net/?f=%5Ctext%7BpH%7D%20%3D%20%5Ctext%7BpK%7D_%7B%5Ctext%7Ba%7D%7D%20%2B%20%5Clog%5Cdfrac%7B%5Ctext%7B%5BA%24%5E%7B-%7D%24%5D%7D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D%5C%5C%5C%5C3.5%20%3D%203.86%20%2B%20%5Clog%5Cdfrac%7B%5Ctext%7B%5BA%24%5E%7B-%7D%24%5D%7D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D%5C%5C%5C%5C%5Clog%5Cdfrac%7B%5Ctext%7B%5BA%24%5E%7B-%7D%24%5D%7D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D%20%3D%203.5%20-%203.86%20%3D%20-0.36%5C%5C%5C%5C%5Cdfrac%7B%5Ctext%7B%5BA%24%5E%7B-%7D%24%5D%7D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D%20%3D%2010%5E%7B-0.36%7D%20%3D%20%5Cmathbf%7B0.44%7D)
He should use a ratio of 0.44 mol lactate to 1 mol of lactic acid.
For example, he could mix equal volumes of 0.044 mol·L⁻¹ lactate and 0.1 mol·L⁻¹ lactic acid.
Answer:
I'm hoping this helps!! it's on quizzes if you're wondering
To find molarity
1) number of mol of solute.
Solute is HCl.
M(HCl)= 1.0+35.5 =36.5 g/mol
25g *1 mol/36.5 g = 25/36.5 mol HCl
2) Molarity is number of mole of the solute in 1 L solution.
150 mL = 0.150 L
(25/36.5 mol HCl )/(0.150 L) = 25/(36.5*0.150) ≈ 4.57≈4.6 mol/L
