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Goryan [66]
3 years ago
12

What discovery was made about particles with an accelerator in 1998

Chemistry
1 answer:
kramer3 years ago
8 0

Answer/Explanation:

In June 1998 in Japan a scientist discovered that neutrinos (which is a type of particle) has weight, mass. This was later proven with some very convincing strong evidence.

<u><em>~ LadyBrain</em></u>

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WILL GIVE BRAINLEST!!
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Coastal flooding occurs when normally dry, low-lying land is flooded by seawater.[1] The extent of coastal flooding is a function of the elevation inland flood waters penetrate which is controlled by the topography of the coastal land exposed to flooding.The seawater can flood the land via from several different paths:
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Lpg contains ? (<br> a. methane (<br> b. butane (<br> c. eyhane (<br> d. none of them
disa [49]
Lpg contains ethane,propane,and butane..hope it helps
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4 years ago
How many grams are in 0.550 mol of cadmium? mass:​
matrenka [14]

Answer:

One mole of cadmium (6multiply1023 atoms) has a mass of 112 grams, as shown in the periodic table on the inside front cover of the textbook. The density of cadmium is 8.65 grams/cm3.

Explanation:

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3 years ago
Which pair of elements would you expect to exhibit the greatest similarity in their physical and chemical properties? Sr, Te Ga,
Anvisha [2.4K]

Answer:

Explanation:

N and P will have similarity

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2,5

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Electronic configuration of  P

1s²2s²2p⁶3s²3p⁵

Electronic configuration bears similarity.

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5 0
3 years ago
1.00 L of a gas at STP is compressed to 473mL. What is the new pressure of gas?
Ratling [72]

Hello!

1.00 L of a gas at STP is compressed to 473 mL. What is the new pressure of gas?

  • <u><em>We have the following data:</em></u>

Vo (initial volume) = 1.00 L  

V (final volume) = 473 mL → 0.473 L  

Po (initial pressure) = 1 atm (pressure exerted by the atmosphere - in STP)  

P (final pressure) = ? (in atm)

  • <u><em>We have an isothermal transformation, that is, its temperature remains constant, if the volume of the gas in the container decreases, so its pressure increases. Applying the data to the equation Boyle-Mariotte, we have:</em></u>

P_0*V_0 = P*V

1*1 = P*0.473

1 = 0.473\:P

0.473\:P = 1

P = \dfrac{1}{0.473}

\boxed{\boxed{P \approx 2.11\:atm}}\:\:\:\:\:\:\bf\green{\checkmark}

<u><em>Answer:  </em></u>

<u><em>The new pressure of the gas is 2.11 atm  </em></u>

___________________________________

\bf\blue{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}

3 0
3 years ago
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