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Triss [41]
3 years ago
8

Why is water classified as a renewable resource?

Chemistry
2 answers:
mrs_skeptik [129]3 years ago
6 0
The answer is a because the water we use it comes back again
777dan777 [17]3 years ago
6 0

Answer:

Option (A)

Explanation:

The resources that can be regenerated again over a specific period of time are known as renewable resources. For example, Wind, Water, and Sunlight.

Water is one of the most necessary and valuable renewable resources. It is considered to be renewable because it can be obtained again and again by undergoing certain processes that form a cycle which is commonly known as the water cycle.

In the water cycle, the water moves from the surface to the atmosphere in the form of vapor by undergoing the process of evaporation and transpiration and comes back to the surface in the form of rainfall, by undergoing the process of condensation.

This is how water changes its phase from one type to another, recycling within the system, thereby regenerating it. So, water is usually classified as a renewable resource.

Thus, the correct answer is option (A).

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An I.V. infusion order is received for Lidocaine 2 g in 250 ml of 0.9% NaCl to be infused at 3 mg/min. What will the flow rate b
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Answer:

The flow rate would be 22.5ml/hr

Explanation:

Volumetric flow rate = Mass flow rate ÷ density

Mass flow rate = 3mg/min = 3mg/min × 60min/1hr = 180mg/hr

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A 101.2 ml sample of 1.00 m naoh is mixed with 50.6 ml of 1.00 m h2so4 in a large styrofoam coffee cup; the cup is fitted with a
Murrr4er [49]

The enthalpy change of the reaction when sodium hydroxide and sulfuric acid react can be calculated using the mass of solution, temperature change, and specific heat of water.

The balanced chemical equation for the reaction can be represented as,

H_{2}SO_{4}(aq) + 2NaOH (aq) ----> Na_{2}SO_{4}(aq) + 2H_{2}O(l)

Given volume of the solution = 101.2 mL + 50.6 mL = 151.8 mL

Heat of the reaction, q = m C .ΔT

m is mass of the solution = 151.8 mL * \frac{1 g}{1 mL} = 151.8 g

C is the specific heat of solution = 4.18 \frac{J}{g. ^{0}C}

ΔT is the temperature change = 31.50^{0}C - 21.45^{0}C = 10.05^{0}C

q = 151.8 g (4.18 \frac{J}{g ^{0}C})(10.05^{0}C) = 6377 J

Moles of NaOH = 101.2 mL * \frac{1L}{1000 mL}*\frac{1.00 mol}{L} = 0.1012 mol NaOH

Moles of H_{2}SO_{4} = 50.6 mL * \frac{1 L}{1000 mL} * \frac{1.0 mol}{1 L} = 0.0506 mol H_{2}SO_{4}

Enthalpy of the reaction = \frac{6377 J*\frac{1kJ}{1000J}}{0.0506 mol} = 126 kJ/mol

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