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Colt1911 [192]
3 years ago
6

The bond angles marked a, b, and c in the molecule below are about __________, __________, and __________, respectively. Select

correct answer. Select correct answer. 120°, 109.5°, 120° 90°, 90°, 90° 109.5°, 120, 109.5° 109.5°, 109.5°, 109.5° 109.5°, 109.5°, 90°
Chemistry
1 answer:
scoundrel [369]3 years ago
6 0

Answer:

109.5°, 120°, 109.5°

Explanation:

The bond angles that are marked as a, b and c in the given molecules is in the order of :

Angle a : 109.5°

Nitrogen is $sp^3$ hybridized and the geometry is tetrahedral. So the bond angle is 109.5°

Angle b : 120°

Carbon is $sp^2$ hybridized, so the shape is trigonal planer and the bond angle is 120°

Angle c : 109.5°

Carbon is $sp^3$ hybridized and the geometry is tetrahedral. The bond angle is 109.5°

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Which of the following statements is true?
VLD [36.1K]

Answer:

C is correct

Explanation:

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The process in which organisms grow and replace worn-out cells is called: A. Cell regeneration B. Cell division C. Mitosis D. Bo
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7 0
4 years ago
the countercurrent arrangement of the two limbs of the loop of henle becomes a multiplier of electrolyte concentration due to wh
enot [183]

The counter-current is the process that occurs in the excretory system. The limbs become the multiplier because of the active transport of the electrolytes that move out.

<h3>What is the loop of Henle?</h3>

The loop of Henle is the part of the excretory system and part of the nephron. It functions in minimizing the water loss in the excretion of urine. There are descending and ascending loops present.

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brainly.com/question/13191804

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5 0
2 years ago
How many grams of carbon dioxide will form if 5.5 g of C3H8 burns in 15 g of O2?
mr Goodwill [35]
C3H8+3O2--->3CO2+8H
Therefore for every 1:3 there are 3 Carbon dioxides that form. That means find the limiting reactant from the two reactants.
5.5g(1mole C3H8/44.03g of C3H8)=0.1249 moled of C3H8 and if for every one C3H8 we can form three CO2. We can assume 0.3747 miles of CO2 will be produced.
15g of O2(1 mole O2/32g of O2)=0.4685moles O2 and if for every three O2 we can produce three CO2 we may assume a 1:1 ratio.
This means C3H8 will be your limiting reactant. Therefore 0.3747 moles of CO2 will be produced.
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5 0
3 years ago
Calculate the molar solubility of ca(io3)2 in each solution below. the ksp of calcium iodate is7.1 × 10−7.
e-lub [12.9K]
In order to find the answer, use an ICE chart:

Ca(IO3)2...Ca2+......IO3- 
<span>some.......0..........0 </span>
<span>less.......+x......+2x </span>
<span>less........x.........2x 
</span>
<span>Ca(IO₃)₂ ⇄ Ca⁺² + 2 IO⁻³
</span>
K sp = [Ca⁺²][IO₃⁻]²
K sp = (x) (2 x)² = 4 x³
7.1 x 10⁻⁷ = 4 x³
<span>x = molar solubility = 5.6 x 10</span>⁻³ M

The answer is 5.6 x 10 ^ 3 M. (molar solubility)
5 0
3 years ago
Read 2 more answers
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