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Vedmedyk [2.9K]
2 years ago
13

Write a simple rule that will tell a person how many water molecule will be lost while putting monosaccharides together to form

polysaccharide ?
Physics
1 answer:
belka [17]2 years ago
3 0
<span>For hydrolysis to monosaccharides, one molecule of a disaccharide needs only one molecule of water. C12H22O11 (sucrose) + H2O = C6H12O6 (glucose) + C6H12O6 (fructose) Structurally, a disaccharide molecule may be viewed as a product formed by the condensation of two molecules of monosaccharides with the elimination of a water molecule. So, only one H2O molecule is needed for the reverse process.</span>
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You might have noticed that a feather falls slowly toward the ground, while a ball falls rapidly. Which statement correctly expl
Brut [27]

Answer:

4. the feather experiences more fluid friction than the ball

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3 years ago
The bearing of lines A and B are 16° 10` and 332° 18`, the value of the included angle BOA is:_______
slega [8]

Answer:

D. 43° 52`  

Explanation:

A bearing is an angle, measured clockwise from the north direction. When solving a bearing problem, it is good to represent the bearings in the given question with diagram.

The diagrammatically representation of the bearing of lines A and B, 16° 10` and 332° 18` respectively given in the question is shown in the figure attached.

At Point A, we will calculate angle ∠BAO.

Calculating the angle ∠BAO

∠BAO = 90° - 16° 10`

          = 73° 50`

At Point B, we will calculate angle ∠ABO.

Calculating the angle ∠ABO

∠ABO = 332° 18` - 270° 0`

          = 62° 18`

At Point O, we will calculate the include angle ∠BOA.

Calculating the angle ∠BOA

∠BAO + ∠ABO + ∠BOA = 180°  (sum of angles in a triangle)

73° 50` + 62° 18` + ∠BOA = 180°

136° 8` + ∠BOA = 180°

∠BOA = 180° - 136° 8`

∠BOA = 43° 52`

The value of the included angle BOA is 43° 52

3 0
3 years ago
In a heat engine if 1000 j of heat enters the system the piston does 500 j of work, what is the final internal energy of the sys
son4ous [18]

Answer:

2500 J

Explanation:

We can solve the problem by using the first law of thermodynamics:

\Delta U =U_f - U_i =Q-W

where

Uf is the final internal energy of the system

Ui is the initial internal energy

Q is the heat added to the system

W is the work done by the system

In this problem, we have:

Q = +1000 J (heat that enters the system)

W = +500 J (work done by the system)

Ui = 2000 J (initial internal energy)

Using these numbers, we can re-arrange the equation to calculate the final internal energy:

U_f = U_i + Q-W=2000 J+1000 J-500 J=2500 J

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3 years ago
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