Atomic Theory. In 1897, English physicist J. J. Thomson (1856–1940) disproved Dalton's idea that atoms are indivisible. ... One of those parts is a negative tiny particle, which Thomson called a corpuscle in 1881.
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Answer:
Balanced equations and mole ratios
In general, mole ratios can be used to convert between amounts of any two substances involved in a chemical reaction.
Explanation:
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This is an exception to the general electronegativity trend. It can be explained by looking at the electron configurations of both elements.
<span>Be:[He]2<span>s2
</span></span><span>B:[He]2<span>s2</span>2<span>p1
</span></span>
When you remove an electron from beryllium, you are taking away an electron from the 2s orbital. When you remove an electron from boron, you are taking an electron from the 2p orbital. The 2p electrons have more energy than the 2s, so it is easier to remove them as they can more strongly resist the effective nuclear charge of the nucleus.
Answer:
The freezing point of the solution is -1.4°C
Explanation:
Freezing point decreases by the addition of a solute to the original solvent, <em>freezing point depression formula is:</em>
ΔT = kf×m×i
<em>Where Kf is freezing point depression constant of the solvent (1.86°C/m), m is molality of the solution (Moles CaBr₂ -solute- / kg water -solvent) and i is Van't Hoff factor.</em>
Molality of the solution is:
-moles CaBr₂ (Molar mass:
189.9g ₓ (1mol / 199.89g) = 0.95 moles
Molality is:
0.95 moles CaBr₂ / 3.75kg water = <em>0.253m</em>
Van't hoff factor represents how many moles of solute are produced after the dissolution of 1 mole of solid solute, for CaBr₂:
CaBr₂(s) → Ca²⁺ + 2Br⁻
3 moles of ions are formed from 1 mole of solid solute, Van't Hoff factor is 3.
Replacing:
ΔT = kf×m×i
ΔT = 1.86°C/m×0.253m×3
ΔT = 1.4°C
The freezing point of water decreases in 1.4°C. As freezing point of water is 0°C,
<h3>The freezing point of the solution is -1.4°C</h3>
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Answer:
The answer to your question is below
Explanation:
To solve this problem use the general formulas for Alkanes, Alkenes, and Alkynes.
Alkanes Cn H(2n + 2)
Alkenes Cn H (2n) "n" is the number of Carbons
Alkynes Cn H (2n - 2)
a) CH₄ C₁ H 2(1) +2 = C₁H₂ + ₂ = CH₄ Alkane
b) C₂H₄ C₂ H₂(₂) Alkene
c) C₃H₆ C(₃) H₂ (₃) Alkene
d) C₂H₂ C₂ H₂(₂) - 2 = C₂H₄ - ₂ = C₂H₂ Alkyne
e) C₁₀H₂₂ C₁₀ H₂(₁₁) Alkene
