All categories

2 years ago

How is sonoluminescence similar to what we see in fireworks?

Chemistry

1 answer:

aalyn [17]2 years ago

7 0

Answer:

Explanation:

Sonoluminescence is a phenomenon in which small gas bubbles burst to produce light when intense sound wave is introduced. Thus, it can be said that sound energy is converted to light energy here. This is similar to the process of how fire crackers work except that sonoluminescence occurs in a liquid medium while firecrackers work in solid media. In firecrackers, when the tip is lit with fire, the initial explosion is caused by the expansion of some nitrogen and carbon present in some constituent compounds, this explosion produces heat which causes some other constituent chemicals in the firecracker to quickly react and produce light. Thus, this can also be said to convert sound energy to light energy and are thus similar to what occurs in sonoluminescence.

You might be interested in

when 6.25 grams of pure iron are allowed to react with oxygen , a black oxide forms. if the product weighs 8.15 g, what is the e

erik [133]

We are given with the mass of pure iron that reacts with oxygen to form an oxide which has a given mass as well. the mass of oxygen reacted is 8.15-6.25 g or 1.9 grams. THen we convert the mass of the reactants to moles. Iron is equal to 0.1119 moles and oxygen is equal to 0.1188. We divide each number to the less amount. Hence iron is 1 and oxygen is approx 1. The empirical formula hence is FeO or ferrous oxide or Iron (II) oxide.

5 0

2 years ago

A chemical reaction takes place inside a flask submerged in a water bath. The water bath contains 8.10kg of water at 33.9 degree

lions [1.4K]

Answer:

The new temperature of the water bath 32.0°C.

Explanation:

Mass of water in water bath ,m= 8.10 kg = 8100 g ( 1kg = 1000g)

Initial temperature of the water = $T_1=33.9^oC=33.9+273K=306.9 K$

Final temperature of the water = $T_2$

Specific heat capacity of water under these conditions = c = 4.18 J/gK

Amount of energy lost by water = -Q = -69.0 kJ = -69.0 × 1000 J

( 1kJ=1000 J)

$Q=m\times c\times \Delta T=m\times c\times (T_2-T_1)$

$-69.0\times 1000 J=8100 g\times 4.18 J/g K\times (T_2-306.9 K)$

$-69,000.0 J=8100 g\times 4.18 J/g K\times (T_2-306.9 K)$

$T_2=304.86 K=304.86 -273^oC=31.86^oC\approx 32.0^oC$

The new temperature of the water bath 32.0°C.

5 0

3 years ago

- How much power does it take to lift a<br> 1,000 N load 10 m in 20 s?

Mariana [72]

Answer:

"500 Joule/sec" is the right answer.

Explanation:

The given values are:

Force,

F = 1000 N

Velocity,

s = 10 m

Time,

t = 20 s

Now,

The power will be:

= $\frac{Force\times Velocity}{Time}$

On putting the values, we get

= $\frac{1000\times 10}{20}$

= $\frac{10000}{20}$

= $500 \ Joule/sec$

5 0

3 years ago

Chose all of the correct statements pure vs mixture

GarryVolchara [31]

Answer:

I think that a pure substance is a form of matter that has a constant composition and properties that are constant throughout the sample. Mixtures are physical combinations of two or more elements and/or compounds.

Explanation:

6 0

3 years ago

Iron and vanadium both have the BCC crystal structure and V forms a substitutional solid solution in Fe for concentrations up to

Bess [88]

Answer:

Explanation:

To find the concentration; let's first compute the average density and the average atomic weight.

For the average density $\rho_{avg}$ ; we have:

$\rho_{avg} = \dfrac{100}{ \dfrac{C_{Fe} }{\rho_{Fe}} + \dfrac{C_v}{\rho_v} }$

The average atomic weight is:

$A_{avg} = \dfrac{100}{ \dfrac{C_{Fe} }{A_{Fe}} + \dfrac{C_v}{A_v} }$

So; in terms of vanadium, the Concentration of iron is:

$C_{Fe} = 100 - C_v$

From a unit cell volume $V_c$

$V_c = \dfrac{n A_{avc}}{\rho_{avc} N_A}$

where;

$N_A$ = number of Avogadro constant.

SO; replacing $V_c$ with $a^3$ ; $\rho_{avg}$ with $\dfrac{100}{ \dfrac{C_{Fe} }{\rho_{Fe}} + \dfrac{C_v}{\rho_v} }$ ; $A_{avg}$ with $\dfrac{100}{ \dfrac{C_{Fe} }{A_{Fe}} + \dfrac{C_v}{A_v} }$ and

$C_{Fe}$ with $100-C_v$

Then:

$a^3 = \dfrac { n \Big (\dfrac{100}{[(100-C_v)/A_{Fe} ] + [C_v/A_v]} \Big) } {N_A\Big (\dfrac{100}{[(100-C_v)/\rho_{Fe} ] + [C_v/\rho_v]} \Big) }$

$a^3 = \dfrac { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100-C_v)A_{v} ] + [C_v/A_Fe]} \Big) } {N_A \Big (\dfrac{100 \times \rho_{Fe} \times \rho_v }{[(100-C_v)/\rho_{v} ] + [C_v \rho_{Fe}]} \Big) }$

$a^3 = \dfrac { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100A_{v}-C_vA_{v}) ] + [C_vA_Fe]} \Big) } {N_A \Big (\dfrac{100 \times \rho_{Fe} \times \rho_v }{[(100\rho_{v} - C_v \rho_{v}) ] + [C_v \rho_{Fe}]} \Big) }$

Replacing the values; we have:

$(0.289 \times 10^{-7} \ cm)^3 = \dfrac{2 \ atoms/unit \ cell}{6.023 \times 10^{23}} \dfrac{ \dfrac{100 (50.94 \g/mol) (55.84(g/mol)} { 100(50.94 \ g/mol) - C_v(50.94 \ g/mol) + C_v (55.84 \ g/mol) } }{ \dfrac{100 (7.84 \ g/cm^3) (6.0 \ g/cm^3 } { 100(6.0 \ g/cm^3) - C_v(6.0 \ g/cm^3) + C_v (7.84 \ g/cm^3) } }$

$2.41 \times 10^{-23} = \dfrac{2}{6.023 \times 10^{23} } \dfrac{ \dfrac{100 *50*55.84}{100*50.94 -50.94 C_v +55.84 C_v} }{\dfrac{100 * 7.84 *6}{600-6C_v +7.84 C_v} }$

$2.41 \times 10^{-23} (\dfrac{4704}{600+1.84 C_v})=3.2 \times 10^{-24} ( \dfrac{284448.96}{5094 +4.9 C_v})$

$\mathbf{C_v = 9.1 \ wt\%}$

4 0

3 years ago

How is sonoluminescence similar to what we see in fireworks?​

How is sonoluminescence similar to what we see in fireworks?