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dezoksy [38]
3 years ago
12

What properties prompted Rutherford to use alpha particles and gold? (What set of experimental conditions would make it easiest

to detect differences in atomic structure?)
Chemistry
1 answer:
Ilia_Sergeevich [38]3 years ago
3 0
<span>Gold is  a stable heavy metal that is easily formed into uniform and thin sheets of measurable property. Alpha particles are relatively massive (compared to electrons) and so deflection is less dramatic and more easily measured. This is the property that prompted Rutherford to use alpha particles and gold. </span>
The particles that are positively charged were fired at the gold foil and when it reached that goldatoms’ nuclei, which are also positively charged, the particles were repelled back
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Write the word and balanced chemical equations for the reaction between:
lesantik [10]

Answer:

Potassium hydroxide + Hydrochloric acid → Potassium chloride + Water.

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Fossil fuels indirectly rely on what source for their energy?
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B, the aún gives enough energy for the fossil fuels.
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What arrangement of electrons would result in a nonpolar molecule
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If the pressure of a 2.00 L sample of gas is 50.0 kPa, what pressure does the gas exert if its volume is decreased to 20.0 mL?
dimulka [17.4K]

Answer:

P₂ = 5000 KPa

Explanation:

Given data:

Initial volume = 2.00 L

Initial pressure = 50.0 KPa

Final volume = 20.0 mL (20/1000=0.02 L)

Final pressure = ?

Solution:

The given problem will be solved through the Boly's law,

"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"

Mathematical expression:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = initial volume

P₂ = final pressure

V₂ = final volume  

Now we will put the values in formula,

P₁V₁ = P₂V₂

50.0 KPa × 2.00L = P₂ × 0.02 L

P₂ = 100 KPa. L/0.02 L

P₂ = 5000 KPa

8 0
2 years ago
Tartaric acid, C4H6O6, has the first ionization constant with the value: Ka1 = 9.20 × 10-4. Calculate the value of pKb for the c
nirvana33 [79]

Answer:

pKb = 10.96

Explanation:

Tartaric acid is a dyprotic acid. It reacts to water like this:

H₂Tart  +  H₂O  ⇄  H₃O⁺   +  HTart⁻         Ka1

HTart⁻  +  H₂O  ⇄  H₃O⁺   +  Tart⁻²           Ka2

When we anaylse the base, we have

Tart⁻²   +  H₂O  ⇄  OH⁻  +  HTart⁻       Kb1

HTart⁻  +  H₂O  ⇄  OH⁻  +    H₂Tart          Kb2

Remember that Ka1 . Kb2  = Kw, plus pKa1 + pKb2 = 14

Kb2 = Kw / Ka1  →   1×10⁻¹⁴ / 9.20×10⁻⁴  = 1.08×10⁻¹¹

so pKb = - log Kb2   → - log 1.08×10⁻¹¹ = 10.96  

7 0
3 years ago
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