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wlad13
1 month ago
8

Classify these compounds as acid, base, salt, or other.? 1. ch3oh 3. hno3 5. nabr

Chemistry
2 answers:
VMariaS [2.9K]1 month ago
7 0

Clarification:

Acids are substances that yield hydrogen ions when they dissociate in water (H^{+}).

For instance, HNO_{3} acts as an acid as it liberates H^{+} ions during dissociation.

            HNO_{3} \rightarrow H^{+} + NO^{-}_{3}

On the other hand, bases release hydroxide ions when dissociating in water (OH^{-}).

For example, NaOH behaves as a base because it produces OH^{-} ions.

Salts result from the combination of positively and negatively charged ions.

For example, NaCl qualifies as a salt.

Thus, the classifications for the compounds provided are as follows.

  • CH_{3}OH - Other
  • HNO_{3} - Acid
  • NaBr - Salt

KiRa [2.9K]1 month ago
5 0
<span> </span><span>1. Other (Alcohol)
3. Acidic
5. Salt

</span>
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eduard [2782]

Answer:

Fatty acids with an even number of carbons, like palmitate, undergo complete β-oxidation in the liver mitochondria, resulting in CO₂, as acetyl-CoA, their end product, can enter the TCA cycle.

On the other hand, odd-number fatty acids such as undecanoic acid generate acetyl-CoA and propionyl-CoA during their final pass. To allow entry into the TCA cycle, propionyl-CoA must go through additional processes, including carboxylation.

The conversion of CO2 and propionyl-CoA into methylmalonyl-CoA is facilitated by propionyl-CoA carboxylase, a biotin-dependent enzyme that is inhibited by avidin. In contrast, the oxidation of palmitate does not require carboxylation.

Explanation:

Fatty acids with an even number of carbons, such as palmitate, are completely oxidized to CO₂ in the liver mitochondria due to the ability of their oxidation product, acetyl-CoA, to enter the TCA cycle where it is further oxidized to CO₂.

Undecanoic acid is classified as an odd-number fatty acid, consisting of 11 carbon atoms. The last stage of β-oxidation for odd-number fatty acids, like undecanoic acid, produces a five-carbon fatty acyl substrate that is oxidized and split into acetyl-CoA and propionyl-CoA. To enter the TCA cycle, propionyl-CoA needs additional reactions such as carboxylation. Since the oxidation occurs using a liver extract, CO₂ must be supplied externally for propionyl-CoA carboxylation, enabling the complete oxidation of undecanoic acid.

The conversion of CO2 and propionyl-CoA into methylmalonyl-CoA is catalyzed by propionyl-CoA carboxylase, which contains biotin. The function of biotin is to activate CO₂ before it is transferred to the propionate group. The addition of avidin obstructs the complete oxidation of undecanoic acid as it binds very tightly to biotin, thereby hindering the activation and transfer of CO₂ to propionate.

In contrast, palmitate oxidation does not require carboxylation, meaning that the presence of avidin doesn't influence its oxidation.

6 0
16 days ago
Commercial grade fuming nitric acid contains about 90.0% HNO3 by mass with a density of 1.50 g/mL, calculate the molarity of the
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x is greater than or equal to 56Step-by-step explanation: He requires at least 56 additional cans. Hence, x should be a minimum of 56.
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12 days ago
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Modern commercial airliners are largely made of aluminum, a light and strong metal. But the fact that aluminum is cheap enough t
lorasvet [2795]

Respuesta:

Un avión fabricado con aluminio puede transportar una mayor cantidad de pasajeros comparado con uno de acero.

Explicación:

La masa total que el avión es capaz de levantar es:

m_{tot}=m_{fuselage}+m_{passangers}

Para el aluminio:

m_{tot}=m_{fus-Al}+m_{pas-Al}

m_{fus-Al}=\delta _{Al}*V_{fuselage}

y

V_{fuselage}=\frac{\pi *L}{4}*[D^2-(D-e)^2]

donde:

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m_{tot}=\delta _{Al}*\frac{\pi *L}{4}*[D^2-(D-e)^2]+m_{pas-Al}

Para el acero (mismo procedimiento):

m_{tot}=\delta _{Steel}*\frac{\pi *L}{4}*[D^2-(D-e)^2]+m_{pas-Steel

Sabiendo que la masa total que el avión puede levantar es constante y que el aluminio tiene una densidad menor que la del acero, podemos afirmar que el avión de aluminio puede levantar un mayor número de pasajeros.

También es posible estimar un peso promedio de los pasajeros para calcular cuántos podría soportar.

5 0
1 month ago
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14 days ago
The average distance between nitrogen and oxygen atoms is 115 pm in a compound called nitric oxide. What is this distance in mil
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Answer:

C) 1.15 × 10⁻⁷ mm

Explanation:

Step 1: Provided information

Average separation between oxygen and nitrogen atoms: 115 pm

Step 2: Change the distance to meters (SI standard unit)

Using the conversion 1 m = 10¹² pm.

115 pm × (1 m/10¹² pm) = 1.15 × 10⁻¹⁰ m

Step 3: Transform the distance to millimeters

Employing the conversion 1 m = 10³ mm.

1.15 × 10⁻¹⁰ m × (10³ mm/1 m) = 1.15 × 10⁻⁷ mm

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