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Harman [31]
3 years ago
7

Please help... I am begging...

Chemistry
1 answer:
iVinArrow [24]3 years ago
5 0

Answer:

The number, such as 98.7 FM, of a radio station represents:

  • <u>the frequency in which is transmitted the radio signal</u>.

Explanation:

<em>The radio FM is the modulated frequency, which means that all the information is sent by just a signal, with different frequencies which difference them</em>, <u>the radio FM use the frequencies from 88 MHz until 108 Mhz (MHz is a measuring unit for the frequency), with a minimal space among them of 0.2 MHz</u>, this last means that you could find a signal in 88.0 MHz, and the next should be 88.2 MHz, next 88.4 MHz and so (at least, regularly the space between two frequencies is more than 0.2 MHz).

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Boyle's law states that the pressure of a gas is always:
natka813 [3]

Answer:

d. inversely proportional to the volume of its container.

Explanation:

Boyle's law  states that at constant temperature and number of moles, the pressure of the gas is inversely proportional to the volume of the gas.

Thus, P ∝ T

P is the pressure

T is the temperature

For two gases at same temperature, the law can be written as:-

{P_1}\times {V_1}={P_2}\times {V_2}

<u>Thus, according to the question, the answer is:- d. inversely proportional to the volume of its container.</u>

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Which body in our solar system has magnetic pole reversals?
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the correct answer would be the sun

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Which molecule has 6 carbon atoms
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Glucose molecule has 6 carbon atoms
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Question 3 of 10
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It’s sulfur because it shows little reactivity.
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Use the following half-reactions to construct a voltaic cell:
velikii [3]

<u>Answer:</u> The correct answer is 3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V

<u>Explanation:</u>

We are given:

Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o=-0.73V\\\\Ag^+(aq.)+e^-\rightarrow Ag(s);E^o=+0.80V

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction. Here, silver will always undergo reduction reaction will get reduced.

Chromium will undergo oxidation reaction and will get oxidized.

The half reactions for the above cell is:

Oxidation half reaction: Cr(s)\rightarrow Cr^{3+}+3e^-;E^o_{Cr^{3+}/Cr}=-0.73V

Reduction half reaction: Ag^{+}+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.80V       ( × 3)

Net equation:  3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.)

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=0.80-(-0.73)=1.53V

Hence, the correct answer is 3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V

4 0
3 years ago
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