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3 years ago

12

What is the role of roughages in the balanced diet

1 answer:

ira [324]3 years ago

5 0

Roughage, also known as Dietary fibre is thought to play an important role in digestive health, heart health, preventing Type 2 diabetes, as well as maintaining energy levels.

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How many moles of glucose (C6H12O6) are in 5.0 liters of a 2.5 M C6H12O6 solution?

miskamm [114]

It's 13 moles I believe

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3 years ago

Explain the type of intermolecular forces exist in glycine molecule.

olga2289 [7]

Answer:

There are three types of intermolecular forces: London dispersion forces (LDF), dipole- dipole interactions, and hydrogen bonding. Molecules can have any mix of these three kinds of intermolecular forces, but all substances at least have LDF.

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3 years ago

The electronic configuration of the al3+ ion is

o-na [289]

Al 1s²2s²2p⁶3s²3p¹ - 3e⁻ → Al³⁺ 1s²2s²2p⁶
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4 years ago

What is the main purpose of a leaf

kenny6666 [7]

Explanation:

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3 years ago

Read 2 more answers

At 400 K, This Reaction Has Kp 8.2 X 104 What Is Kp At 400 K For The Following Reaction

kondor19780726 [428]

The given question is incomplete, here is a complete question.

At 400 K, this Reaction has $K_p=8.2\times 10^{-4}$

$SO_3(g)\rightleftharpoons SO_2(g)+\frac{1}{2}O_2(g)$

What Is $K_p$ at 400 K for the following reaction?

$2SO_3(g)\rightleftharpoons 2SO_2(g)+O_2(g)$

(A) 8.2 x 10⁻⁴

(B) 2.9 x 10⁻²

(C) 6.7 x 10⁻⁷

(D) 1.6 x 10⁻⁷

Answer : The correct option is, (C) $6.7\times 10^{-7}$

Explanation :

The given chemical equation follows:

$SO_3(g)\rightleftharpoons SO_2(g)+\frac{1}{2}O_2(g)$

The equilibrium constant for the above equation, $K_p=8.2\times 10^{-4}$ .

We need to calculate the equilibrium constant for the following equation of above chemical equation, which is:

$2SO_3(g)\rightleftharpoons 2SO_2(g)+O_2(g)$ , $K_p'$

The equilibrium constant for the doubled reaction will be the square of the initial reaction.

Or, we can say that

If the equation is multiplied by a factor of '2', the equilibrium constant will be the square of the equilibrium constant of initial reaction.

The value of equilibrium constant for the following reaction is:

$K_{p}'=(K_p)^2$

$K_{p}'=(8.2\times 10^{-4})^2$

$K_{p}'=6.7\times 10^{-7}$

Hence, the value of equilibrium constant for the following reaction is, $6.7\times 10^{-7}$

7 0

3 years ago