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Salsk061 [2.6K]
3 years ago
5

Organic acids and bases in their ___________ form are soluble in organic solvents, but the corresponding _________ are more wate

r-soluble. Therefore, converting acids
and bases to the corresponding _________ provides a convenient tool to separate acidic, alkaline and neutral organic compounds.
Chemistry
1 answer:
gizmo_the_mogwai [7]3 years ago
3 0

Answer:

a. neutral

b. salts

c. salt

Explanation:

Organic salts are a dense number of ionic compounds with innumerable characteristics. They are previously derived from an organic compound, which has undergone a transformation that allows it to be a carrier of a charge, and that in addition, its chemical identity depends on the associated ion.

Organic salts are usually stronger acids or bases than inorganic salts. This is because, for example, in the amine salts, it has a positive charge due to its bond with an additional hydrogen: A + -H. Then, in contact with a base, donate the proton to be a neutral compound again

RA + H + B => RA + HB

H belongs to A, but it is written as it is involved in the neutralization reaction.

On the other hand, RA + can be a large molecule, unable to form solids with a crystalline network stable enough with the hydroxyl anion or oxyhydrile OH–.

When this is so, salt RA + OH– behaves as a strong base; even as basic as NaOH or KOH

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Explanation:

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3 years ago
Which of the following is an example of a neutralization reaction?
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HCl (aq) + NaOH (aq)→H2O (l) + NaCl (aq)
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A projectile is fired with speed v0 at an angle theta from the horizontal from the horizontal as shown in the figure.
irga5000 [103]

Answer:

v₀ = √(2gH/(sin²θ)) = (sin θ)√(2gH)

v₀ = √(gR/(sin2θ))

Explanation:

An image of the artillery officer, the hill and path of motionof the projectile is attached to this solution.

Given, R, H, g and θ (theta)

Using the equations of motion, we can get the initial velocity v₀

First of, we need to resolve this motion into the vertical and horizontal axis.

The horizontal component of the initial velocity, v₀ₓ = v₀ cos θ

Vertical component of the initial velocity, v₀ᵧ = v₀ sin θ

When the projectile reaches maximum height, Velocity at max height, vₕ = 0m/s

From equations of motion,

vₕ = v₀ᵧ - gt

0 = v₀ sinθ - gt

t = v₀ sinθ/g

This is the time taken to reach maximum height. The time take to comolete the toyal flight, T = 2t = (2v₀ sinθ)/g

The maximum height to be reached, H can be calculated from the equations of motion too

H = vₕt - 0.5gt² = 0 - 0.5g((v₀ sinθ)/g)²

H = (0.5g v₀² sin²θ)/g²

H = (v₀² sin²θ)/2g

The range, or horizontal distance to be covered by the projectile, R, will be calculated using the horizontal component of the initial Velocity, v₀ₓ = v₀ cos θ, this horizontal velocity is constant all through the motion, so, no acceleration in the horizontal direction.

R = v₀ₓT =  (v₀ cos θ)((2v₀ sinθ)/g)

R = (v₀²(2cosθsinθ)/g)

2cosθsinθ = sin2θ

R = v₀²(sin2θ)/g

So, writing v₀ in terms of all the other parameters,

v₀ = √(2gH/(sin²θ)) =  (sinθ)√(2gH

v₀ = √(gR/(sin2θ))

4 0
2 years ago
2. Given the unbalanced equation:
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2 Na + 2 H2O → 2 NaOH + H2 (balanced equation)
The answer would be 2, since 2 in the coefficient of both Na and NaOH
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The volume of 0.6305 kg of a liquid is 0.430 L. What is its density in g/mL
motikmotik

Answer: 1.466 g/mL

Explanation:

\frac{0.6305 \text{kg}}{0.430 \text{L}} \cdot \frac{1000 \text{g}}{1 \text{kg}} \cdot  \frac{1 \text{L}}{1000 \text{mL}} = 1.466 \,\frac{\text{g}}{\text{mL}}

6 0
1 year ago
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