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jolli1 [7]
4 years ago
5

Y^2- 4x² - 4y—8x-16=0transverse axis ​

Chemistry
1 answer:
Nostrana [21]4 years ago
7 0

Answer:

  • line segment of length 8 between the vertices at (-1, -2) and (-1, 6)

Explanation:

The equation can be rearranged to standard form.

  (y^2 -4y) -4(x^2 +2x) = 16

  (y^2 -4y +4) -4(x^2 +2x +1) = 16 +4 -4

  (y -2)^2 -4(x +1)^2 = 16

  (y -2)^2 /16 -(x +1)^2/4 = 1

This is of the form ...

  (y -k)^2/a^2 -(x -h)^2/b^2 = 1

where the transverse axis is 2a and the center is (h, k). Here, a=4, so 2a = 8.

The transverse axis is a vertical line segment of length 8, centered on (-1, 2).

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Answer:

Their vector sum is zero, so CO2 therefore has no net dipole. (b) In H2O, the O–H bond dipoles are also equal in magnitude, but they are oriented at 104.5° to each other. Hence the vector sum is not zero, and H2O has a net dipole moment.

❤✔

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3 years ago
Given the reactant side of the total ionic equation for the neutralization reaction of lithium hydroxide (LiOH) with hydrochlori
Sunny_sXe [5.5K]
<span>Answer:
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</span><span>
</span><span>Li⁺ (aq) + OH⁻ (aq) + H⁺ (aq) + Cl⁻(aq) → Li⁺ (aq) + Cl⁻ (aq) + H₂O(l)</span><span />

<span>Explanation:
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<span>1) Combine the cation Li⁺ (aq) with the anion Cl- (aq) to form LiCl(s).
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<span>LiCl is a solid soluble substance, a typical ionic compound. So, it will reamain as separate ions in the product side: Li⁺ + CL⁻</span>

<span>2) Combine the anion OH⁻ with the cation H⁺ to form H₂O(l).
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<span>Since, the ionization of H₂O is low, it will remain as liquid in the product side: H₂O(l)</span>

<span>3) Finally, you can wirte the total ionic equation:
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Li⁺ (aq) + OH⁻ (aq) + H⁺ (aq) + Cl⁻(aq) → Li⁺ (aq) + Cl⁻ (aq) + H₂O(l)



8 0
3 years ago
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nexus9112 [7]
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