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marissa [1.9K]
3 years ago
9

If a satellite or spaceship explodes in outer space, you can see the explosion from a distance but you cannot hear it. Why?

Chemistry
2 answers:
lidiya [134]3 years ago
7 0

Answer:The correct answer is option C.

Explanation:

Mechanical wave is a wave that produce from the oscillation of matter and transfer energy  through the medium.They are incapable of moving through vacuum.

Electromagnetic wave is a wave composed of two vectors vibrating perpendicular to each other in the same direction in which wave is moving.The two vectors are electric field and magnetic field.They are capable of travelling through vacuum.

Light wave is an electromagnetic wave.Electromagnetic waves can easily travel through space. Where as sound is a mechanical wave which requires medium to travel.

Hence, correct answer is option C.

labwork [276]3 years ago
3 0
<span>C) Sound is a mechanical wave and needs a medium to travel through while light is an electromagnetic wave. 

There is no air for the sound waves to travel, therefore, no sound waves being transmitted. </span>
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PLEASE HELP ME!!! What does this diagram represent? A. Uniformitarianism B. Fossil Correlation C. Faunal Succession D. The water
Leni [432]
That would be, A. Uniformitarianism
3 0
3 years ago
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Describe how to use a transfer pipet to deliver 10.00 mL of liquid by placing the steps in the correct order.
nexus9112 [7]

This question is incomplete, the complete question is;

Describe how to use a transfer pipet to deliver 10.00 mL of liquid by placing the following steps in the correct order;

- Replace the bulb with your index finger and wipe excess liquid off the outside of the pipette.

- Place the tip of the pipette against the side of the beaker and drain the liquid from the pipette until the bottom of the meniscus reaches the calibration mark.

- Transfer the pipette to the receiving vessel.

- Drain the pipette by gravity while holding the tip against the side of the receiving vessel.

- Use a rubber bulb to suck liquid up past the 10.00 ml calibration mark.

- Suck up a third volume of liquid past the 10.00 ml calibration mark.

- Discard the first two pipette volumes of the liquid to rinse the pipette.

Answer:

- Use a rubber bulb to suck liquid up past the 10.00 ml calibration mark.

- Discard the first two pipette volumes of the liquid to rinse the pipette.

- Suck up a third volume of liquid past the 10.00 ml calibration mark.

- Replace the bulb with your index finger and wipe excess liquid off the outside of the pipette.

- Place the tip of the pipette against the side of the beaker and drain the liquid from the pipette until the bottom of the meniscus reaches the calibration mark.

- Transfer the pipette to the receiving vessel.

- Drain the pipette by gravity while holding the tip against the side of the receiving vessel.  

Explanation:  

First of all, We use a suction device to suck up liquid into pipet. Then we will fill the pipet up to the 10 mL marks and will discard the initial two volumes of liquid and will take a final third volume. We will replace the bulb with index finger and will drain it by placing the tip of pipet at the wall of beaker and drain the liquid.

Arranged in the following steps correctly;

- Use a rubber bulb to suck liquid up past the 10.00 ml calibration mark.

- Discard the first two pipette volumes of the liquid to rinse the pipette.

- Suck up a third volume of liquid past the 10.00 ml calibration mark.

- Replace the bulb with your index finger and wipe excess liquid off the outside of the pipette.

- Place the tip of the pipette against the side of the beaker and drain the liquid from the pipette until the bottom of the meniscus reaches the calibration mark.

- Transfer the pipette to the receiving vessel.

- Drain the pipette by gravity while holding the tip against the side of the receiving vessel.

5 0
2 years ago
I need help again...
Mandarinka [93]
The theory of evolution
3 0
2 years ago
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A compound with a molar mass of 60g/mol is 40.4% carbon, 6.7% hydrogen and 53.3% oxygen (by mass). determine the emperical and m
Fittoniya [83]

<span>A compound is found to be 40.0% carbon, 6.7% hydrogen and 53.5% oxygen. Its molecular mass is 60. g/mol. 
</span>Q1)
Empirical formula is the simplest ratio of whole numbers of components making up a compound.
the percentages have been given, therefore we can calculate for 100 g of the compound.

                                C                            H                        O
Mass in 100 g      40.0 g                       6.7 g                   53.5 g
Molar mass            12 g/mol                1 g/mol                 16 g/mol
Number of moles   40.0/12= 3.33         6.7/1 = 6.7          53.5/16 = 3.34
Divide by the least number of moles  
                             3.33/3.33 = 1           6.7/3.33 = 2.01   3.34/3.33 = 1.00
after rounding off
C - 1 
H - 2
O - 1

Empirical formula - CH₂O

Q2)
Molecular formula is the actual number of components making up the compound.
To find the number of empirical units we have to find the mass of one empirical unit.
Mass of one empirical unit = CH₂O - 12 + (1x2) + 16 = 30 g
Mass of one mole of compound = 60 g
Number of empirical units = 60 g / 30 g = 2
Therefore molecular formula - 2(CH₂O) 
 Molecular formula - C₂H₄O₂
4 0
2 years ago
Be sure to answer all parts. Calculate the molality, molarity, and mole fraction of FeCl3 in a 24.0 mass % aqueous solution (d =
Anna71 [15]

Answer:

m= 1.84 m

M= 1.79 M

mole fraction (X) =

Xsolute= 0.032

Xsolvent = 0.967

Explanation:

1. Find the grams of FeCl3 in the solution: when we have a mass % we assume that there is 100 g of solution so 24% means 24 g of FeCl3 in the solution. The rest 76 g are water.

2. For molality we have the formula m= moles of solute / Kg solvent

so first we pass the grams of FeCl3 to moles of FeCl3:

24 g of FeCl3x(1 mol FeCl3/162.2 g FeCl3) = 0.14 moles FeCl3

If we had 76 g of water we convert it to Kg:

76 g water x(1 Kg of water/1000 g of water) = 0.076 Kg of water

now we divide m = 0.14 moles FeCl3/0.076 Kg of water

m= 1.84 m

3. For molarity we have the formula M= moles of solute /L of solution

the moles we already have 0.14 moles FeCl3

the (L) of solution we need to use the density of the solution to find the volume value. For this purpose we have: 100 g of solution and the density d= 1.280 g/mL

The density formula is d = (m) mass/(V) volume if we clear the unknown value that is the volume we have that (V) volume = m/d

so V = 100 g / 1.280 g/mL = 78.12 mL = 0.078 L

We replace the values in the M formula

M= 0.14 moles of FeCl3/0.078 L

M= 1.79 M

3. Finally the mole fraction (x)  has the formula

X(solute) = moles of solute /moles of solution

X(solvent) moles of solvent /moles of solution

X(solute) + X(solvent) = 1

we need to find the moles of the solvent and we add the moles of the solute like this we have the moles of the solution:

76 g of water x(1 mol of water /18 g of water) = 4.2 moles of water

moles of solution = 0.14 moles of FeCl3 + 4.2 moles of water = 4.34 moles of solution

X(solute) = 0.14 moles of FeCl3/4.34 moles of solution = 0.032

1 - X(solute) = 1 - 0.032 = 0.967

6 0
2 years ago
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