Answer:
14 OH⁻(aq) + 2 Cr³⁺(aq) = Cr₂O₇²⁻(aq) + 7 H₂O(l) + 6 e⁻
Explanation:
In order to balance a half-reaction we use the ion-electron method.
Step 1: Write the half-reaction
Cr³⁺(aq) = Cr₂O₇²⁻(aq)
Step 2: Perform the mass balance, adding H₂O(l) and OH⁻(aq) where appropriate
14 OH⁻(aq) + 2 Cr³⁺(aq) = Cr₂O₇²⁻(aq) + 7 H₂O(l)
Step 3: Perform the electric balance, adding electrons where appropriate
14 OH⁻(aq) + 2 Cr³⁺(aq) = Cr₂O₇²⁻(aq) + 7 H₂O(l) + 6 e⁻
<em>★</em><em> </em><em>«</em><em> </em><em><u>Synthetic reactions in Citric Acid Cycle</u></em><em><u> </u></em><em>»</em><em> </em><em>★</em>
- <em>The overall reaction for the citric acid cycle is as follows: acetyl-CoA + 3 NAD+ + FAD + GDP + P + 2H2O = CoA-SH + 3NADH + FADH2 + 3H+ + GTP + 2CO2. </em>
- <em>Many molecules in the citric acid cycle serve as key precursors for other molecules needed by cells.</em>
<em>hope</em><em> </em><em>it</em><em> is</em>
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Answer:
Add to a 500mL volumetric flask 300mL of water, the 129mL of the 11.6M HCl solution and then complete to volume with water
Explanation:
To make 500mL = 0.500L of a 3M HCl from the 11.6M HCl stock we need first to find the moles of HCl we need:
<em>Moles HCl:</em>
0.500L * (3mol / L) = 1.5 moles of HCl are needed
These moles are obtained from the 11.6M HCl solution. The volume required is:
1.5mol * (1L / 11.6moles HCl) = 0.129L = 129mL must be added to the solution.
That means to prepare the 500mL of the 3M HCl you need to:
Add to a 500mL volumetric flask 300mL of water, the 129mL of the 11.6M HCl solution and then complete to volume with water