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3 years ago

8

Drag each tile to the correct box a chemical reaction takes place in which energy is released arrange the reactions characterist

ics in order from start to finish

1 answer:

wlad13 [49]3 years ago

8 0

Where are the tiles my wise sir

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A circuit consists of one 330 ohm resistor in series with two other resistors (470 ohms and 220 ohms) connected in parallel. Wha

umka21 [38]

Answer:

power drain on an ideal battery, P = 0.017 W

Given:

$R_{1} = 330\ohm$

$R_{2} = 470\ohm$

$R_{3} = 220\ohm$

Since, $R_{2} = 470\ohm$ and $R_{3} = 220\ohm$ are in parallel and this combination is in series with $R_{1} = 330\ohm$ , so,

Equivalent resistance of the circuit is given by:

$R_{eq} = \frac{R_{2}R_{3}}{R_{2} + R_{3}} + R_{1}$

$R_{eq} = \frac{470\times 220}{470 + 220} + 330$

$R_{eq} = 149.85 + 330 = 479.85 \ohm$

power drain on an ideal battery, P = $\frac{V^{2}}{R_{eq}}$

P = $\frac{2.9^{2}}{479.85}$

P = 0.017 W

4 0

4 years ago

Give one example of a thermodynamic cycle that does not account for the carnot efficiency.

Arturiano [62]

Thermo-Electrochemical converter (UTEC) is a thermodynamic cycle that does not account for the Carnot Efficiency.

The Carnot cycle is a hypothetical cycle that takes no account of entropy generation. It is assumed that the heat source and heat sink have perfect heat transfer. The working fluid also remains in the same phase, as opposed to the Rankine cycle, in which the fluid changes phase. A practical thermodynamic cycle, such as the Rankine cycle, would achieve at most 50% of the Carnot cycle efficiency under similar heat source and heat sink temperatures.

<h3>What is Thermo-Electrochemical converter?</h3>

In a two-cell structure, a thermo-electrochemical converter converts potential energy difference during hydrogen oxidation and reduction to heat energy.

It employs the Ericsson cycle, which is less efficient than the Carnot cycle. In a closed system, it converts heat to electrical energy. There are no external input or output devices.

This means there will be no mechanical work to be done, as well as no exhaust. As a result, Carnot efficiency is not taken into account in this cycle. Carnot efficiency is accounted for by other options such as turbine and engine.

Learn more about Thermo-Electrochemical converter here:

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4 0

2 years ago

Consider the following examples of homeostatic regulation: In response to an increase in plasma K concentrations, secretion of t

TiliK225 [7]

Answer:

Both are examples of negative feedback regulation.

Explanation:

The maintenance of the homeostasis in the body is controlled by the the feedback regulation of the body. Two main types of feedback regulation are positive regulation and negative regulation.

The negative regulation occurs when the final product of the reactions inhibits the further secretion of that product. In the given examples of aldosterone and calcium mechanism, the secretion of aldosterone and calcium decreases as the normal levels are acheived in the body.

Thus, the answer is both are examples of negative feedback regulation.

5 0

3 years ago

To enhance the effective surface, and hence the chemical reaction rate, catalytic surfaces often take the form of porous solids.

marysya [2.9K]

Answer:

See solution with all the conditions considered. A gaseous mixture of A and B for which species A is chemically consumed at the catalytic surface.

The total pore reaction rate is stated below and it can be inferred by applying the bellow analogy.

4 0

3 years ago

How many diffraction maxima are contained in a region of the Fraunhofer single-slit pattern, subtending an angle of 2.12°, for a

Luba_88 [7]

Answer:

6

Explanation:

We are given that

$\theta=2.12^{\circ}$

Slid width,a=0.110 mm= $0.11\times 10^{-3} m$

$1mm=10^{-3} m$

Wavelength, $\lambda=582 nm=582\times 10^{-9}$ m

$1nm=10^{-9} m$

We have to find the number of diffraction maxima are contained in a region of the Fraunhofer single-slit pattern.

$asin\theta=\frac{2N+1}{2}\lambda$

Using the formula

$0.11\times 10^{-3}sin(2.12)=\frac{2N+1}{2}(582\times 10^{-9})$

$2N+1=\frac{0.11\times 10^{-3}sin(2.12)\times 2}{582\times 10^{-9}}$

$2N+1=13.98$

$2N=13.98-1=12.98$

$N=\frac{12.98}{2}\approx 6$

Hence, 6 diffraction maxima are contained in a region of the Fraunhofer single-slit pattern

5 0

4 years ago