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Lunna [17]
3 years ago
12

Is a metal baking tray a conductor or insulator or a radiator

Chemistry
2 answers:
Travka [436]3 years ago
8 0
I would think conductor since it absorbs heat and which is radiation but that's not the question and the heat it absorbs it touches the food which is conduction
Svetllana [295]3 years ago
5 0
The tray is a radiator, b/c it helps the heat transfer/spread throughout the oven while baking something for example. Unlike a conductor, having energy pass through the tray too quickly barely getting any heat directly around the tray. And an Insulator preventing the heat from getting through the metal baking tray at all.     

Hope this helps!  <span> ^ ω ^  </span>
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Crossword please help must finish in 15 mins
kogti [31]

Across:

10. The circulatory system transfers nutrients, gases, liquids, and heat around the body.

11. The circulatory system transports heat, which helps regulate temperature.

13. The place where oxygen enters the blood and carbon dioxide leaves the blood.  Lungs.

15. A gas that is transported in arteries from the lungs to the rest of the body via the heart.  Oxygen.

Down:

2. The heart, blood, and vessels.  Circulatory System.

4. Blood in arteries is bright red because it is rich in oxygen.

6. A waste gas that is transported in veins from the body to the lungs via the heart.  Carbon di Oxide.

<u>Explanation:</u>

The circulatory system includes blood vessels, blood, and heart. This system provides the body tissues with oxygen and some nutrients. It also carries hormones and eliminates needless waste products.

This transportation takes place between the cells via blood throughout the body. The channel that blood passes through is a blood vessel that is pumped by an organ called heart. The heart directs the blood passing all over the body.

The lungs are a duo of air-filled, spongy organs positioned on both sides of a human's chest. Its main function is to take in air present in the atmosphere and transfer oxygen to the bloodstream. From where it gets circulated throughout the body.

6 0
3 years ago
The vapor pressure of ethanol at 25 Celsius is 0.0773 atm. Calculate the vapor pressure in mmHg and torr. Round each of your ans
Elden [556K]

Answer:

hope it helps!Brainliest pls

Explanation:

8 0
2 years ago
Look at Table 4 in the procedure portion of the experiment. Calculate the pH you would expect each of the buffer solutions (A, B
ivann1987 [24]

The pH of the buffer solutions as determined using the Henderson–Hasselbalch equation are:

  • A. pH = 4.75
  • B. pH = 4.05
  • C. pH = 3.75
  • D. pH = 5.75
  • E. pH = 5.45

<h3>What is the pH of the solutions?</h3>

The pH of a buffer is determined using the Henderson–Hasselbalch equation shown below:

  • pH = pKₐ + log([A⁻]/[HA])

A. Volume of acetic acid = 5 mL; Volume of sodium acetate = 5 mL; pka of acetic acid = 4.75

The solutions of acetic acid and sodium acetate are equimolar;

pH = 4.75 + log(1)

pH = 4.75

B. Volume of acetic acid = 5 ml; Volume of sodium acetate = 1 mL; pka of acetic acid = 4.75

The solutions of acetic acid and sodium acetate are equimolar;

pH = 4.75 + log(1/5)

pH = 4.05

C. Volume of acetic acid = 10 ml; Volume of sodium acetate = 1 mL; pka of acetic acid = 4.75

The solutions of acetic acid and sodium acetate are equimolar;

pH = 4.75 + log(1/10)

pH = 3.75

D. Volume of acetic acid = 1 ml; Volume of sodium acetate = 10 mL; pka of acetic acid = 4.75

The solutions of acetic acid and sodium acetate are equimolar;

pH = 4.75 + log(10/1)

pH = 5.75

E. Volume of acetic acid = 1 ml; Volume of sodium acetate = 5 mL; pka of acetic acid = 4.75

The solutions of acetic acid and sodium acetate are equimolar;

pH = 4.75 + log(5/1)

pH = 5.45

In conclusion, the pH of the buffer solutions are determined using the Henderson–Hasselbalch equation.

Learn more about buffers at: brainly.com/question/22390063

#SPJ1

4 0
2 years ago
What is the molarity of a 5 L solution that contains 7.5 moles of NaOH?
Aneli [31]

Answer:1.5 M

Explanation:

5 0
2 years ago
Consider a voltaic cell where the anode half-reaction is Zn(s) → Zn2+(aq) + 2 e− and the cathode half-reaction is Sn2+(aq) + 2 e
Sonbull [250]

Answer:

Option (B) is correct

Explanation:

Oxidation: Zn\rightarrow Zn^{2+}+2e^{-}

Reduction: Sn^{2+}+2e^{-}\rightarrow Sn

---------------------------------------------------------------------------

Overall: Zn+Sn^{2+}\rightarrow Zn^{2+}+Sn

Nernst equation for this cell reaction at 25^{0}\textrm{C}:

E_{cell}=(E_{Sn^{2+}\mid Sn}^{0}-E_{Zn^{2+}\mid Zn}^{0})-\frac{0.059}{n}log\frac{[Zn^{2+}]}{[Sn^{2+}]}

Where, n is number of electron exchanged and species inside third bracket represents concentrations in molarity.

Here, n = 2, E_{cell}=0.660V and [Zn^{2+}]=2.5\times 10^{-3}M

So, plug in all the given values into above equation:

0.660V=(-0.136V+0.76V)-\frac{0.059}{2}log\frac{2.5\times 10^{-3}M}{[Sn^{2+}]}

So, [Sn^{2+}]=4.2\times 10^{-2}M

As the value "0.059" varies from literature to literature and 4.2\times 10^{-2}M is most closest to 3.3\times 10^{-2}M therefore option (B) is correct.

8 0
3 years ago
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