Which of the following has helped preserve resources while cutting the demand for energy?

Chemistry

2 answers:

sweet [91]2 years ago

4 0

A is the correct answer

Mekhanik [1.2K]2 years ago

3 0

Answer: Option (D) is the correct answer.

Explanation:

When there is minimum use of energy and we are still able to preserve resources then it means we are cutting the demand for energy.

Therefore, when we are treating water with chlorine then it means without any electricity or energy we are cleaning the water. But at the same time chlorinated water is harmful to use unless chlorine is dissolved in limiting quantity.

Whereas energy efficient appliances will use the electricity efficiently.

Landfill will require the use of machines or man power therefore, energy will be used.

On the other hand, when rules are framed against the use of chlorofluorocarbons then there will be lesser depletion of ozone layer. Hence, use of energy will be minimum.

Thus, we can conclude that legislation against the use of chlorofluorocarbons has helped preserve resources while cutting the demand for energy.

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An emulsifying agent is typically characterized by having ____. a. one polar end c. two polar ends b. one nonpolar end d. one po

tangare [24]

Answer:

An emulsifying agent is typically characterized by having d. one polar end and one nonpolar end.

Explanation:

Emulsifiers are substances that have the ability to bind, for example, fats with those substances that have mostly water in their conformation. In other words, the emulsifier facilitates mixtures of two or more immiscible liquid substances.

This is because the molecules of an emulsifier are often lipophilic (attract oil) at one end and hydrophilic (attract water) at the other. In other words it consists of a polar (hydrophilic) head group and a non-polar (hydrophobic) tail.

An emulsifying agent is typically characterized by having d. one polar end and one nonpolar end.

4 0

3 years ago

How much heat energy, in kilojoules, is required to convert 72.0 gg of ice at −−18.0 ∘C∘C to water at 25.0 ∘C∘C ? Express your a

Ghella [55]

Answer: The enthalpy change is 34.3 kJ

Explanation:

The conversions involved in this process are :

$(1):H_2O(s)(-18^0C)\rightarrow H_2O(s)(0^0C)\\\\(2):H_2O(s)(0^0C)\rightarrow H_2O(l)(0^0C)\\\\(3):H_2O(l)(0^0C)\rightarrow H_2O(l)(25^0C)$

Now we have to calculate the enthalpy change.

$\Delta H=[m\times c_{s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{l}\times (T_{final}-T_{initial})]$

where,

$\Delta H$ = enthalpy change = ?

m = mass of water = 72.0 g

$c_{s}$ = specific heat of ice = $2.09J/g^0C$

$c_{l}$ = specific heat of liquid water = $4.184J/g^0C$

n = number of moles of water = $\frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{72.0g}{18g/mole}=4.00moles$

$\Delta H_{fusion}$ = enthalpy change for fusion = 6010 J/mole

Now put all the given values in the above expression, we get

$\Delta H=[72.0g\times 2.09J/g^0C\times (0-(-18)^0C]+4.00mole\times 6010J/mole+[72.0g\times 4.184J/g^)C\times (25-0)^0C]$ $\Delta H=34279.8J=34.3kJ$ (1 KJ = 1000 J)