Answer: Oxygen and glucose are both reactants in the process of cellular respiration.
Explanation: Oxygen and glucose are both reactants in the process of cellular respiration. The main product of cellular respiration is ATP; waste products include carbon dioxide and water.
Answer:
(3R,4R)-4-bromohexan-3-ol
Explanation:
In this case, we have reaction called <u>halohydrin formation</u>. This is a <u>markovnikov reaction</u> with <u>anti configuration</u>. Therefore the halogen in this case "Br" and the "OH" must have <u>different configurations</u>. Additionally, in this molecule both carbons have the <u>same substitution</u>, so the "OH" can go in any carbon.
Finally, in the product we will have <u>chiral carbons</u>, so we have to find the absolute configuration for each carbon. On carbon 3 we will have an "R" configuration on carbon 4 we will have also an "R" configuration. (See figure 1)
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The methyl orange would be the best indicator for titrating the weak base sodium bicarbonate using HCl titrant
When weak base is titrated with strong acid then , then solution is slightly acidic at end point . If weak acid is titrated with strong base then , the solution is slightly basic because salt formed will hydrolyzed to a certain extent .
In acid base titration at the end point the amount of the acid becomes chemically equivalent to the amount of the base present .The methyl orange would be the best indicator for titrating the weak base with strong acid .
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Step 1: write the equation:
P₄(s) + 6F₂(g) → 4PF₃(g)
Step 2: Molar mass of P₄ = 30.97 g/mol × 4 = 123.88 g/mol
Step 3: Number of moles of phosphorus
n = m/M
n = 8.5 g/123.88g/mol
n = 0.07 moles
Step 4: 0.07 × 12 = 0.84 moles of fluorine.
Fluorine is diatomic gas so we multiplied the number of moles by 12.
Step 5: To find the mass of fluorine we multiply the number of moles with the molar mass.
Mass of fluorine = 0.84 × 228
= 191.52 grams.
Answer:
Cathode: Ag
Anode: Br₂
Explanation:
In the cathode must occur a reduction, so it's more likely to a metal atom be in the cathode. For the metals given the reduction reactions and the potential of reduction are:
Ag⁺ + e⁻ ⇒ Ag⁰ E° = + 0.80 V
Fe⁺² + 2e⁻ ⇒ Fe⁰ E° = - 0.44 V
Al⁺³ + 3e⁻ ⇒ Al⁰ E° = -1.66 V
As the potential for Ag is the higher, the reduction will occur for it first, so in the cathode will produce Ag.
For the anode an oxidation must occurs, so the reactions for the nonmetals are:
F₂ + 2e⁻ ⇒ 2F⁻ E° = +2.87 V
Cl₂ + 2e⁻ ⇒ 2Cl⁻ E° = +1.36 V
Br₂ + 2e⁻ ⇒ 2Br⁻ E° = +1.07 V
For oxidation, the less the E°, the faster the reaction will occur, so Br₂ will be formed in the anode.
