Answer:
The correct answer is True.
Explanation:
In the kidney, the handle of Henle is a fork-like tube located in the nephrons. It is the portion of the nephron that leads from the proximal contoured tubule to the distal contoured tubule. The primary filtrate from the glomerulus passes into the proximal contoured tubule, which connects to the descending branch of the loop of Henle, which has low permeability to ions and urea (solutes) but is very permeable to water.
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D i think if not here’s this a collection of beliefs or practices mistakenly regarded as being based on scientific method
Its is true that light from an unshaded bulb radiates in all directions
2MnO4-(aq) + 16H+ + 10Cl-(aq) ⇔ 2Mn2+(aq) + 8H2O + 5Cl2 (g)
<h3>Further explanation</h3>
Given
Reaction(unbalanced)
Mno4- (aq) + Cl- (aq) → Mn2+ + Cl2 (g)
Required
Half reaction
Solution
1. Add coefficient(equalizing atoms in reaction)
2. Adding H₂O on the O-deficient side.
3. Adding H⁺ on the H-deficient side.
4. Adding e⁻
5. Equalizing the number of electrons and sum the all the half-reaction
1.MnO₄⁻(aq) = Mn²⁺(aq) reduction
2.MnO4(aq) = Mn2+(aq) + 4H2O
3. MnO4-(aq) + 8H+ = Mn2+(aq) + 4H2O
4. MnO4-(aq) + 8H+ + 5e- = Mn2+(aq) + 4H2O
Cl⁻(aq) = Cl₂(g) oxidation
1. 2Cl-(aq) = Cl2 (g)
2-3 none
4. 2Cl-(aq) = Cl2 (g) + 2e-
5.
MnO4-(aq) + 8H+ + 5e- = Mn2+(aq) + 4H2O x2
2Cl-(aq) = Cl2 (g) + 2e- x5
2MnO4-(aq) + 16H+ + 10e- = 2Mn2+(aq) + 8H2O
10Cl-(aq) = 5Cl2 (g) + 10e-
<em>2MnO4-(aq) + 16H+ + 10Cl-(aq) ⇔ 2Mn2+(aq) + 8H2O + 5Cl2 (g)</em>