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3 years ago

What is the balanced equation for the combustion of magnesium?

Chemistry

1 answer:

Lostsunrise [7]3 years ago

6 0

Answer:

2Mg + O2 → 2MgO

Explanation:

In all conbustion you should know, that reactans are an specific compound and O2, so the products must be CO2 and H2O, or in this case, the corresponding oxide.

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Who believed and tested his idea that microorganisms are carried through the air?

kondor19780726 [428]

Louis Pasteur designed a procedure to test whether sterile nutrient broth could spontaneously generate microbial life. To do this, he set up two experiments.

Hope this helps !

Have a great day! :)

7 0

3 years ago

Read 2 more answers

What is Displacement reaction . Explain with an example.

jok3333 [9.3K]

Explanation:

A displacement reaction is the one wherein the atom or a set of atoms is displaced by another atom in a molecule. For instance, when iron is added to a copper sulphate solution, it displaces the copper metal. A + B-C → A-C + B.

6 0

3 years ago

Describe the relationship between frequency and wavelength in your own words

Montano1993 [528]

Are frequency and wavelength the same thing? No, they are not the same but each is mathematically related to the other. Effectively, the wavelength is the distance between one wave peak and the next wave peak, or in other words, the distance between one wave high point and the next high point. Alternatively it could of course be said that wavelength is the distance between one wave low point and the next wave low point, but lets not get pedantic about it.

Think of waves in the ocean where a person may be observing the top of one wave and the top of the next wave. The wavelength is the distance between these two wave tops, or peaks. With waves in the ocean, the frequency of the waves will be the number of times that a wave peak crosses any given point on the ocean. It is probably easiest to measure frequency of waves from the ocean by standing on the beach and counting how many waves come up on the sand relative to any given time frame. Frequency is typically measured in how many waves per second but with ocean waves we are better to measure how many waves per minute because naturally the frequency will be less than one per second.

There is actually quite a bit of science over how ocean waves travel around our planet because high and low tide in the ocean are created by the moon. There is a theory that the moon creates waves that have a wavelength equal to half of the circumference of planet Earth. This is because there is a high tide at Earths point that is closest to the moon and then another high point that is at the greatest distance from the moon. The problem is that to keep up with the moon one needs to travel around the Earths equator at about 1800 kph, which is impractical for an ocean wave because they quite simply cannot travel at that fast a speed or velocity (technically angular velocity). This is what causes ocean waves to become so messy at times.

When discussing waves, most people are most comfortable discussing electromagnetic waves because almost all communication systems relied on in modern society are based on these waves and their frequency. When collecting electromagnetic waves, like TV waves, for example, there are several components to the antenna. One of them will typically have a loop of metal, which is where the required energy waves (carrying the required signal) are picked up or collected by the antenna. Don’t worry about the other components of the antenna because most are there simply to remove unwanted background waves that may spoil the quality of the signal that we collect. With the TV antenna, the distance across this “collection loop” is the wavelength that the antenna is tuned to collect.

The reason for a loop on the piece of metal that collects TV waves, rather than using a straight piece of wire, is so that all wavelengths that are close to the required one, will be collected. To get slightly more technical, in modern systems we have “frequency modulation”, which is what FM stand for. This means we deliberately make minor adjustments to the precise frequency, but I better not go into that.

The frequency of a TV wave that is being collected is the number of times in any time frame, that a wave front or wave peak, will cross the collection point. With typical electromagnetic waves like TV waves, we use a frequency that is measured to be so many Hertz. The Hertz is the standard measure of such things and it is equal to a number of wavelengths per second. The reason for this is that electromagnetic waves travel at the speed of light, which is incredibly fast.

If we are talking in the old fashioned “long wave” AM radio waves, then the wavelength is often several hundred metres in length. In Melbourne, Australia, the nearest large city to my home, the government owned ABC used a frequency of 774 kHz for many decades. They still do in fact, although most people tuned in probably rely on a repeater station these days and these will broadcast in a higher frequency. 774 kHz is a frequency of 774 thousand cycles per second. This sounds like a high frequency when compared to most other waves, even sound waves, yet because radio waves travel so fast, the wavelength is slightly greater than 387 metres in wavelength, which is almost 424 yards in wavelength.

When comparing wavelength to frequency, one is the inverse of the other so this means that the higher the frequency, the shorter the wavelength, with the rate of travel (velocity) being the factor that determines what sort of figures we come up with when comparing one to the other.
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3 years ago

Calculate the EMF between copper and silver Ag+e-E=0.89v<br>Cu=E=0.34v

bogdanovich [222]

Answer:

Depending on the $E^\circ$ value of $\rm Ag^{+} + e^{-} \to Ag\; (s)$ , the cell potential would be:

$0.55\; \rm V$ , using data from this particular question; or
approximately $0.46\; \rm V$ , using data from the CRC handbooks.

Explanation:

In this galvanic cell, the following two reactions are going on:

The conversion between $\rm Ag\; (s)$ and $\rm Ag^{+}$ ions, $\rm Ag^{+} + e^{-} \rightleftharpoons Ag\; (s)$ , and
The conversion between $\rm Cu\; (s)$ and $\rm Cu^{2+}$ ions, $\rm Cu^{2+}\; (aq) + 2\, e^{-} \rightleftharpoons \rm Cu\; (s)$ .

Note that the standard reduction potential of $\rm Ag^{+}$ ions to $\rm Ag\; (s)$ is higher than that of $\rm Cu^{2+}$ ions to $\rm Cu\; (s)$ . Alternatively, consider the fact that in the metal activity series, copper is more reactive than silver. Either way, the reaction is this cell will be spontaneous (and will generate a positive EMF) only if $\rm Ag^{+}$ ions are reduced while $\rm Cu\; (s)$ is oxidized.

Therefore:

The reduction reaction at the cathode will be: $\rm Ag^{+} + e^{-} \to Ag\; (s)$ . The standard cell potential of this reaction (according to this question) is $E(\text{cathode}) = 0.89\; \rm V$ . According to the 2012 CRC handbook, that value will be approximately $0.79\; \rm V$ .

The oxidation at the anode will be: $\rm Cu\; (s) \to \rm Cu^{2+} + 2\, e^{-}$ . According to this question, this reaction in the opposite direction ( $\rm Cu^{2+}\; (aq) + 2\, e^{-} \rightleftharpoons \rm Cu\; (s)$ ) has an electrode potential of $0.34\; \rm V$ . When that reaction is inverted, the electrode potential will also be inverted. Therefore, $E(\text{anode}) = -0.34\; \rm V$ .

The cell potential is the sum of the electrode potentials at the cathode and at the anode:

$\begin{aligned}E(\text{cell}) &= E(\text{cathode}) + E(\text{anode}) \\ &= 0.89 \; \rm V + (-0.34\; \rm V) = 0.55\; \rm V\end{aligned}$ .

Using data from the 1985 and 2012 CRC Handbook:

$\begin{aligned}E(\text{cell}) &= E(\text{cathode}) + E(\text{anode}) \\ &\approx 0.7996 \; \rm V + (-0.337\; \rm V) \approx 0.46\; \rm V\end{aligned}$ .

5 0

3 years ago

Describe how changing the particles changed the atom

Zanzabum

Bonds between atoms break and new ones form and energy is either given out or taken in

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3 years ago