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3 years ago

Foods rich in simple carbohydrates or starch but low in fat or fiber tend to

Chemistry

2 answers:

Alex Ar [27]3 years ago

7 0

Answer is: A. be digested and absorbed quickly.

Starch (carbohydrate) is a polymeric carbohydrate consisting of a large number of glucose units bonded by glycosidic bond.

Starch is a white, tasteless and odorless powder that is insoluble in cold water or alcohol.

Food rich in carbohydrates has high energy content, because human body can digest it very fast.

Some fruits have a lot of cellulose (also carbohydrate) and fiber, so human body cannot digest it or digest it very slowly.

GREYUIT [131]3 years ago

5 0

A because if it was high in fiber it would go slowly that's why people eat allot of fiber a day before a race or workout

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The Galapagos Islands formed over a hot spot. The majority of the volcanoes here are ________________ volcanoes. A) shield B) ci

Diano4ka-milaya [45]

I think the answer would be shield volcanoes.

5 0

2 years ago

1. What would you need to know in order to predict what color of skin your offspring would have?

Blizzard [7]

Answer:

I believe you would need to know how much melanon you have and how much melanon your partner has. Correct me if i'm wrong. I hope this helps!

6 0

2 years ago

What is the mass of 0.7891 mol of (fe2o3)?

Serjik [45]

Molar mass F₂O₃ = 159.69 g/mol

1 mole ------------------ 159.69 g
0.7891 moles ---------- ? g

mass = 0.7891 x 159.69 / 1

= 126.01 g

hope this helps!

4 0

2 years ago

The dissociation of molecular iodine into iodine atoms is represented as i2(g) ⇌ 2i(g) at 1000 k, the equilibrium constant kc fo

ki77a [65]

First, we must write the balanced chemical equation for the process:

I₂ (g) ⇌ 2I (g)

The chemical reactions that occur in a closed container can reach a state of chemical equilibrium that is characterized because the concentrations of the reactants and products remain constant over time. The equilibrium constant of a chemical reaction is the value of its reaction quotient in chemical equilibrium.

The equilibrium constant (Kc) is expressed as the ratio between the molar concentrations (mol/L) of reactants and products. Its value in a chemical reaction depends on the temperature, so it must always be specified.

So, the equilibrium constant for the above reaction is,

Kc = $\frac{[I]^{2} }{[I_{2} ]}$

The initial concentration of I₂ is:

[I₂]₀ = $\frac{0.0456 mol}{2.28 L}$ → [I₂]₀ = 0.0200 M

To facilitate the visualization of the reaction, we can write it in the following way,

I₂ ⇌ 2I

0.0200 M - initial

0.0200 M - x 2x equilibrium

As the reaction doesn't go to completion, at equilibrium an amount x dissociates of the initial concentration of I₂ to produce 2x of I (because of the stequiometric coefficients).

At equilibrium [I₂] = 0.0200 M - x and [I] = 2x. By substituting this in the above equation for kc we will be able to find the value of x and then calculate the concentrations of both gases in equilibrium.

If kc = 3.80ₓ10⁻⁵ we have,

Kc = $\frac{[I]^{2} }{[I_{2} ]} = \frac{(2x)^{2} }{0.0200 - x} = 3.80x10^{-5}$

Since the dissociation constant has such a low value, we can assume that the degree of dissociation of I₂ is very low, therefore x ≈ 0 and,

$\frac{(2x)^{2} }{0.0200 - x} = \frac{4x^{2} }{0.0200} = 3.80x10^{-5}$

→ x = 4.36ₓ10⁻⁴ M

Then, the concentrations of the gases I₂ and I at equilibrium are,

[I₂] = 0.0200 M - x = 0.0200 M - 4.36ₓ10⁻⁴ M → [I₂] = 0.0196 M

[I] = 2x = 2 x 4.36ₓ10⁻⁴ M → [I] = 8.72ₓ10⁻⁴ M

3 0

2 years ago

HELP PLEASE ASAP

grin007 [14]

Agreed with guy above thanks

6 0

2 years ago