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Gekata [30.6K]
2 years ago
14

Write a 150 word paragraph or two that describes at least three everyday things that exist or occur because of science. Make sur

e you use examples different from those given in the lesson.
Chemistry
2 answers:
netineya [11]2 years ago
8 0
<span>Specular reflection, or the act of seeing ones self in the mirror, is an everyday part of science that we commonly use without much thought about it. This is basically a reflection of light off a reflective surface, the image does not scattered but is printed on the surface for you to see. Though quantum mechanics may explain how a magnet works scientist really have no true understanding of why charged particles create magnetic fields when they are moved around. The microwave oven is another form of everyday thing that casually uses science. The science behind it is called electromagnetic energy. Basically it operates on a micro frequency that passes nearly harmlessly through plastics, glass, ceramics, etc to stimulate the atoms in liquids and meats to a reaction point that cooks them. Of course these micro waves also over stimulate metals thus why metal in the microwave is very bad.</span>
Radda [10]2 years ago
7 0

Practically any process you can think of in this modern world is made possible and available because of some form of scientific discovery. For example, I am able to take online classes, like I am doing right now, on the internet because of the scientific discoveries that have been made in order to create the laptops, the internet, and the online software used for the schooling. In addition to that has been previously stated, daily medicine or vitamin supplements that I take every day are a result of science and scientific discoveries that have been done to prove they help keep people healthy. Another thing we do every day thanks to science is washing our hands with soap every day and after we use the bathroom. We do this because of our understanding of germs that have been formed thanks to the process of the scientific method and scientific discoveries.

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A transition in the balmer series for hydrogen has an observed wavelength of 434 nm. Use the Rydberg equation below to find the
DanielleElmas [232]

Answer:

i. n = 5

ii. ΔE = 7.61 × 10^{-46} KJ/mole

Explanation:

1. ΔE = (1/λ) = -2.178 × 10^{-18}(\frac{1}{n^{2}_{final} } - \frac{1}{n^{2}_{initial}  })

    (1/434 × 10^{-9}) = -2.178 × 10^{-18} (\frac{n^{2}_{initial} - n^{2}_{final}  }{n^{2}_{final} n^{2}_{initial}   })

⇒ 434 × 10^{-9} = (1/-2.178 × 10^{-18})\frac{n^{2}_{final} *n^{2}_{initial}   }{n^{2}_{initial} - n^{2}_{final}    }

But, n_{final} = 2

434 × 10^{-9} = (1/2.178 × 10^{-18})\frac{2^{2} n^{2}_{initial}  }{n^{2}_{initial} - 2^{2}  }

434 × 10^{-9}  × 2.178 × 10^{-18} = (\frac{4n^{2}_{initial}  }{n^{2}_{initial} - 4 })

⇒ n_{initial} = 5

Therefore, the initial energy level where transition occurred is from 5.

2. ΔE = hf

     = (hc) ÷ λ

    = (6.626 × 10−34 × 3.0 × 10^{8} ) ÷ (434 × 10^{-9})

    = (1.9878 × 10^{-25}) ÷ (434 × 10^{-9})

    = 4.58 × 10^{-19} J

    = 4.58 × 10^{-22} KJ

But 1 mole = 6.02×10^{23}, then;

energy in KJ/mole = (4.58 × 10^{-22} KJ) ÷ (6.02×10^{23})

         = 7.61 × 10^{-46} KJ/mole

7 0
2 years ago
24 how many moles are in 2.04 × 1024 molecules of h2o?
Sergeeva-Olga [200]
The answer is 3.39 mol.

<span>Avogadro's number is the number of molecules in 1 mol of substance.
</span><span>6.02 × 10²³ molecules per 1 mol.
</span>2.04 × 10²⁴<span> molecules per x.

</span>6.02 × 10²³ molecules : 1 mol = 2.04 × 10²⁴ molecules : x
x = 2.04 × 10²⁴ molecules * 1 mol : 6.02 × 10²³ molecules
x = 2.04/ 6.02 × 10²⁴⁻²³ mol
x = 0.339 × 10 mol
<span>x = 3.39 mol
</span>
8 0
2 years ago
How many milligrams of MgI2 must be added to 257.7 mL of 0.087 M KI to produce a solution with [I−] = 0.1000 M?
lbvjy [14]

Answer:

The answer is 465.6 mg of MgI₂ to be added.

Explanation:

We find the mole of ion I⁻ in the final solution

C = n/V -> n = C x V = 0.2577 (L) x 0.1 (mol/L) = 0.02577 mol

But in the initial solution, there was 0.087 M KI, which can be converted into mole same as above calculation, equal to 0.02242 mol.

So we need to add an addition amount of 0.02577 - 0.02242 = 0.00335 mol of I⁻. But each molecule of MgI₂ yields two ions of I⁻, so we need to divide 0.00335 by 2 to find the mole of MgI₂, which then is 0.001675 mol.

Hence, the weight of MgI₂ must be added is

Weight of MgI₂ = 0.001675 mol x 278 g/mol = 0.4656 g = 465.6 mg

4 0
2 years ago
At a temperature of __________ °c, 0.444 mol of co gas occupies 11.8 l at 889 torr.
Novay_Z [31]
<span>ideal gas law is: PV = nRT P = pressure (torr) = 889 torr V = volume (Liters) = 11.8 L n = moles of gas = 0.444 mol R = gas constant = 62.4 (L * torr / mol * k) solve for T (in kelvin) T = PV/nR T = (889*11.8)/(.444*62.4) T = 378.6 K convert to C (subtract 273) T = 105.6 deg C</span>
3 0
2 years ago
Read 2 more answers
How much water(in grams) at its boiling point can be vaporized by adding 1.50 kJ of heat? The molar heat of vaporization for wat
LekaFEV [45]

Answer:

0.66g of water

Explanation:

Molar heat of vaporization of any substance is defined as the heat necessary to vaporize 1 mole of the substance.

If heat of vaporization of water is 40.79kJ/mol and you add 1.50kJ, the moles you vaporize are:

1.50kJ × (1mol / 40.79kJ) = 0.0368 moles of water.

As molar mass of water is 18.01g/mol, mass of water that can be vaporized are:

0.0368 moles × (18.01g / mol) = <em>0.66g of water</em>

6 0
2 years ago
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