What are the building block of lipids?

What is the largest contributor to the greenhouse effect?

Sloan [31]

Answer:

burning of fossil fuels

Explanation:

7 0

3 years ago

Explain the difference between heat and temperature. Does 1 L of water at 658°F have more, less, or the same quantity of energy

pogonyaev

The answer is- The energy of 1 L water at temperature 347.78 °C have more energy as 1 L of water at temperature 65°C.

Heat is a type of energy that causes a person's body to feel hot or cold.

While the temperature of an object is a parameter that indicates how hot or cold the object is.

How is the temperature in degree Fahrenheit converted to degree celsius?

To convert the temperature in Fahrenheit to Celsius, subtract 32 and multiply by 5/9.

° $C =( ^0F -32) * \frac{5}{9}$

Now, heat is a form of energy that flows from hotter object to colder object and temperature indicates whether the object is hot or cold by measuring its average kinetic energy.
Now, the given temperature of 1 L water is 658 °F. This temperature in degree celsius is calculated as-

°C $= (658 F-32) *\frac{5}{9} = 347.78 \° C$

Now, higher the temperature, higher is the energy of water. Thus, the energy of 1 L water at 347.78 °C have more energy as 1 L of water at 65°C.

To learn more about heat and temperature, visit:

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3 0

2 years ago

A 225-g sample of aluminum was heated to 125.5 oc, then placed into 500.0 g water at 22.5 oc. (the specific heat of aluminum is

Sidana [21]

when heat gained = heat lost

when AL is lost heat and water gain heat

∴ (M*C*ΔT)AL = (M*C*ΔT) water

when M(Al) is the mass of Al= 225g

C(Al) is the specific heat of Al = 0.9

ΔT(Al) = (125.5 - Tf)

and Mw is mass of water = 500g

Cw is the specific heat of water = 4.81

ΔT = (Tf - 22.5)

so by substitution:

∴225* 0.9 * ( 125.5 - Tf) = 500 * 4.81 * (Tf-22.5)

∴Tf = 30.5 °C

7 0

3 years ago

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Be sure to answer all parts. Calculate the pH during the titration of 30.00 mL of 0.1000 M KOH with 0.1000 M HBr solution after

Fantom [35]

Answer:

(a) pH = 12.73

(b) pH = 10.52

(c) pH = 1.93

Explanation:

The net balanced reaction equation is:

KOH + HBr ⇒ H₂O + KBr

The amount of KOH present is:

n = CV = (0.1000 molL⁻¹)(30.00 mL) = 3.000 mmol

(a) The amount of HBr added in 9.00 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(9.00 mL) = 0.900 mmol

This amount of HBr will neutralize an equivalent amount of KOH (0.900 mmol), leaving the following amount of KOH:

(3.000 mmol) - (0.900 mmol) = 2.100 mmol KOH

After the addition of HBr, the volume of the KOH solution is 39.00 mL. The concentration of KOH is calculated as follows:

C = n/V = (2.100 mmol) / (39.00 mL) = 0.0538461 M KOH

The pOH and pH of the solution can then be calculated:

pOH = -log[OH⁻] = -log(0.0538461) = 1.2688

pH = 14 - pOH = 14 - 1.2688 = 12.73

(b) The amount of HBr added in 29.80 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(29.80 mL) = 2.980 mmol

This amount of HBr will neutralize an equivalent amount of KOH, leaving the following amount of KOH:

(3.000 mmol) - (2.980 mmol) = 0.0200 mmol KOH

After the addition of HBr, the volume of the KOH solution is 59.80 mL. The concentration of KOH is calculated as follows:

C = n/V = (0.0200 mmol) / (59.80 mL) = 0.0003344 M KOH

The pOH and pH of the solution can then be calculated:

pOH = -log[OH⁻] = -log(0.0003344) = 3.476

pH = 14 - pOH = 14 - 3.476 = 10.52

(c) The amount of HBr added in 38.00 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(38.00 mL) = 3.800 mmol

This amount of HBr will neutralize all of the KOH present. The amount of HBr in excess is:

(3.800 mmol) - (3.000 mmol) = 0.800 mmol HBr

After the addition of HBr, the volume of the analyte solution is 68.00 mL. The concentration of HBr is calculated as follows:

C = n/V = (0.800 mmol) / (68.00 mL) = 0.01176 M HBr

The pH of the solution can then be calculated:

pH = -log[H⁺] = -log(0.01176) = 1.93

4 0

3 years ago

Which process in the light-dependent reactions results in the release of hydrogen ions, electrons, and oxygen?

vodomira [7]

The high-energy electron travels down an electron transport chain, losing energy as it goes.
Some of the released energy drives pumping of \text H^+H+H, start superscript, plus, end superscript ions from the stroma into the thylakoid interior, building a gradient.
H^+H+H, start superscript, plus, end superscript ions from the splitting of water also add to the gradient.
 H^+H+H, start superscript, plus, end superscript ions flow down their gradient and into the stroma, they pass through ATP synthase, driving ATP production in a process known as chemiosmosis.

3 0

3 years ago

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