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pochemuha
3 years ago
13

Suppose you are a particle og watet in a lake. Describe what happens to you when a motorboat passes by.

Chemistry
1 answer:
Tamiku [17]3 years ago
3 0

i will move and collapse with other h20 particles and splash around


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Rutherford’s gold foil experiment gave evidence that an atom is mostly empty space. true false
Sholpan [36]

The given statement is true .

<h3>What is Rutherford’s gold foil  experiment?</h3>
  • A piece of gold foil was hit with alpha particles, which have a favorable charge. Most alpha particles went right around. This showed that the gold particles were mostly space.
  • The Rutherford gold leaf investigation supposed that most (99%) of all the mass of an atom is in the middle of the atom, that the nucleus is very small (105 times small than the length of the atom) and that is positively captured.
  • For the distribution experiment, Rutherford enjoyed a metal sheet that could be as thin as practicable. Gold is the most malleable of all known metals. It can easily be converted into very thin sheets. Hence, Rutherford established a gold foil for his alpha-ray scattering experimentation.

To learn more about Rutherford’s gold foil experiment, refer to:

brainly.com/question/4113533

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4 0
1 year ago
Calculate the density of a liquid if 58.9 ml of it has a mass of 46.08 g. answer in units of g/ml.
alexira [117]
Density of a solution is mass of solution per unit volume
Density = mass/volume
mass of solution is 46.08 g
volume of solution is 58.9 mL 
since mass and volume is known, density can be calculated
density = 46.08 g / 58.9 mL = 0.78 g/mL 
8 0
3 years ago
Suppose a mercury thermometer contains 3.250g of mercury and has a capillary that is 0.180mm in diameter.. How far does the merc
Helga [31]
1) You need to get volume of both temperatures by using first attached formula V= Mass/DensityV_1 = (3.250 g)/(13.596 g/(cm^3)) = .2390 cm^3&#10;At 25 degrees C (V_2e have:&#10;&#10;V_2  = (3.250 g)/(13.534 g/(cm^3)) = .2401 cm^3

2) Using the second formula you get the height of 0 degree 
h = V/(pi * r^2)&#10;&#10;&#10;h_1 =(.2390 cm^3)/(pi*(.009 cm)^2)  = 939.2 cm

(radius in cm is (0.180 mm) / 2 * (1 cm)/(10 mm) or .009 cm

3) Then with h1 you can easily get the height of 25 degrees 
h_2 =(.2401 cm^3)/(pi*(.009 cm)^2) = 943.5 cm

 Subtract 943.5 cm - 939.2 cm, and obtain a rise in mercury height of 4.3 cm

5 0
3 years ago
Read 2 more answers
To draw the Lewis structure of the polyatomic ion,CIO3 you would have to _____ those in the structures of Cl, O, O, and O. Selec
NikAS [45]
D. Add one electron to, 
3 0
3 years ago
Na2S2O3 + AgBr NaBr + Na3ſAg(S203)2] What is
gtnhenbr [62]

Answer:

Mass of NaBr produced  = 23.67 g

Explanation:

Given data:

Mass of AgBr = 42.7 g

Mass of NaBr produced = ?

Solution:

Chemical equation:

2Na₂S₂O₃ + AgBr    →    NaBr + Na₃(Ag(S₂O₃)₂

Number of moles of AgBr:

Number of moles = mass/molar mass

Number of moles = 42.7 g/ 187.7 g/mol

Number of moles = 0.23 mol

now we will compare the moles of AgBr with NaBr.

             AgBr        :         NaBr

                1            :           1

              0.23       :         0.23

Mass of NaBr:

Mass = number of moles × molar mass

Mass = 0.23 mol × 102.89 g/mol

Mass = 23.67 g

8 0
3 years ago
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