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Alecsey [184]
2 years ago
6

An aqueous solution is made from a molecular compound rather than from an ionic compound. If some ions are present in solution,

which best describes the solute? a weak electrolyte a nonelectrolyte a strong electrolyte a covalent electrolyte
Chemistry
2 answers:
brilliants [131]2 years ago
6 0

Answer:

A

Explanation:

I dont know why some people make it so difficult thats all it is

Blizzard [7]2 years ago
4 0

Answer:

As a result, electrolyte solutions readily conduct electricity. ... By contrast, if a compound dissociates to a small extent, the solution will be a weak conductor of electricity; ... Typically, nonelectrolytes are primarily held together by covalent rather than ionic bonds. ... Explain why some molecules do not dissolve in water.

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Which has a higher percentage of the diequatorial-substituted conformer compared with the diaxialsubstituted conformer: trans-1,
kherson [118]

Answer:

cis-1-tert-butyl-3-methylcyclohexane will have a higher percentage of the diequatorial-substituted conformer when compared with the diaxialsubstituted conformer.

Explanation:

The two compound contain or have high stability with the substituent group is at equatorial position but the tert-betyl group in cis-1-tert-butyl-3-methylcyclohexane is larger than the methyl group in trans-1,4-dimethylcyclohexane.

Thus, the equatorial position will be more favorable for the substituent group in the cis-1-tert-butyl-3-methylcyclohexane, therefore having higher percentage of the diequatorial substituted conformer compared with that of diaxial-substituted conformer.

3 0
3 years ago
All of the following are uses of a lake/pond except:
umka2103 [35]
Well give me the options lol
6 0
2 years ago
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.

3 0
3 years ago
How many moles of solute are present in 2 L of a 25M potassium nitrate (KNO3) solution?
Maksim231197 [3]
Mole= Molarity. Volume(L) = (25)(2)= 50 moles
8 0
3 years ago
The equation for the combustion of CH4 (the main component of natural gas) is
Lera25 [3.4K]

Heat produced =  -13588.956 kJ

<h3>Further explanation</h3>

Given

The reaction of combustion of Methane

CH4(g)+2O2(g)→CO2(g)+2H2O(g) ΔH∘rxn=−802.3kJ

271 g of CH4

Required

Heat produced

Solution

mol of 271 g CH₄ (MW=16 g/mol0

mol = mass : MW

mol = 271 : 16

mol = 16.9375

So Heat produced :

= mol x ΔH°rxn

= 16.9375 mol x −802.3kJ/mol = -13588.956 kJ

6 0
2 years ago
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