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Citrus2011 [14]
4 years ago
12

Which molecule is amphoteric? A) NH4+ B )H2O C )H2CO3− D )H2SO4

Chemistry
2 answers:
fiasKO [112]4 years ago
5 0

Answer: B ) H_2O]

Explanation:- According to the Bronsted Lowry conjugate acid-base theory, an acid is defined as a substance which donates protons and a base is defined as a substance which accepts protons.

Amphoteric substances are substances which can act as both acids and bases and thus can accept and donate H^+ ions.

H_2O\rightarrow H^++OH^_

Here water acts as an acid as it donates H^+ ions.

H_2O+H^+\rightarrow H_3O^+

Here water acts as a base as it accepts H^+ ions.

Maslowich4 years ago
4 0
The answer is letter choice B



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A 100 gram glass container contains 200 grams of water and 50.0 grams of ice all at 0°c. a 200 gram piece of lead at 100°c is ad
ASHA 777 [7]

0 \; \textdegree{\text{C}}

Explanation:

Assuming that the final (equilibrium) temperature of the system is above the melting point of ice, such that all ice in the container melts in this process thus

  • E(\text{fusion}) = m(\text{ice}) \cdot L_{f}(\text{water}) = 66.74 \; \text{kJ} and
  • m(\text{water, final}) = m(\text{water, initial}) + m(\text{ice, initial}) = 0.250 \; \text{kg}

Let the final temperature of the system be t \; \textdegree{\text{C}}. Thus \Delta T (\text{water}) = \Delta T (\text{beaker}) = t(\text{initial})  - t_{0} = t \; \textdegree{\text{C}}

  • Q(\text{water}) &= &c(\text{water}) \cdot m(\text{water, final}) \cdot \Delta T (\text{water})= 1.047 \cdot t\; \text{kJ} (converted to kilojoules)
  • Q(\text{container}) &= &c(\text{glass}) \cdot m(\text{container}) \cdot \Delta T (\text{container})= 0.0837 \cdot t \; \text{kJ}
  • Q(\text{lead}) &= &c(\text{lead}) \cdot m(\text{lead}) \cdot \Delta T (\text{lead})= 0.0255 \cdot (100 - t)\; \text{kJ}

The fact that energy within this system (assuming proper insulation) conserves allows for the construction of an equation about variable t.

E(\text{absorbed} ) = E(\text{released})

  • E(\text{absorbed} ) = E(\text{fushion}) + Q(\text{water}) + Q(\text{container})
  • E(\text{released}) =  Q(\text{lead})

Confirm the uniformity of units, equate the two expressions and solve for t:

66.74 + 1.047 \cdot t + 0.0837 \cdot t = 0.0255 \cdot (80 - t)

t \approx -55.95\; \textdegree{\text{C}} < 0\; \textdegree{\text{C}} which goes against the initial assumption. Implying that the final temperature does <em>not</em> go above the melting point of water- i.e., t \le 0 \; \textdegree{\text{C}}. However, there's no way for the temperature of the system to go below 0 \; \textdegree{\text{C}}; doing so would require the removal of heat from the system which isn't possible under the given circumstance; the ice-water mixture experiences an addition of heat as the hot block of lead was added to the system.

The temperature of the system therefore remains at 0 \; \textdegree{\text{C}}; the only macroscopic change in this process is expected to be observed as a slight variation in the ratio between the mass of liquid water and that of the ice in this system.

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Answer:

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