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3 years ago

5

What substance helps suspend fat in a watery digestive mixture, making fat more available to digestive enzymes?

1 answer:

TEA [102]3 years ago

7 0

I believe that the answer to the question provided above is that the <span>substance helps suspend fat in a watery digestive mixture, making fat more available to digestive enzymes is known as starch.</span>
Hope my answer would be a great help for you. If you have more questions feel free to ask here at Brainly.

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When part of earths surface is tilted towards the sun what does it cause

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Summer Solstice is what happens when part of the Earth's surface is tilted towards the sun..

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3 years ago

Read 2 more answers

Select the correct answer. Two charged objects, A and B, exert an electric force on each other. What happens if the distance bet

VikaD [51]

Answer:

The electric force between them decreases

Explanation:

The force between two charged particle is given by :

$F=\dfrac{kq_1q_2}{r^2}$

Where

r is the distance between charges

If the distance between the charges is increased, the electric force gets decreased as there is an inverse relation between force and distance.

Hence, the correct option is (c) "The electric force between them decreases"

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2 years ago

Define duplet and octet??

Bogdan [553]

Answer:
duplet-A set of two things or people
Octet-A group of 8 things or people

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3 years ago

A container is at a pressure of 3 atm and a temperature of 280K. What is the new temperature when the pressure is reduced to 1.5

GenaCL600 [577]

Answer:

140 K

Explanation:

Step 1: Given data

Initial pressure of the gas (P₁): 3 atm

Initial temperature of the gas (T₁): 280 K

Final pressure of the gas (P₂): 1.5 atm

Final temperature of the gas (T₂): ?

Step 2: Calculate the final temperature of the gas

We have a gas whose pressure is reduced. If we assume an ideal behavior, we can calculate the final temperature of the gas using Gay-Lussac's law.

T₁/P₁ = T₂/P₂

T₂ = T₁ × P₂/P₁

T₂ = 280 K × 1.5 atm/3 atm = 140 K

6 0

3 years ago

4. A student started with a 0.032 g sample of copper which he took through the series of reactions described in this experiment.

Elodia [21]

Answer:

$Y=48.6\%$

Explanation:

Hello,

In this case, we can consider the following chemical reaction for the oxidation of copper which only occurs at high temperatures:

$2Cu+O_2\rightarrow 2CuO$

In such a way, for 0.032 grams of copper, the following grams of copper (II) oxide (black product) are yielded:

$m_{CuO}=0.032gCu*\frac{1molCu}{63.546gCu} *\frac{2molCuO}{2molCu}*\frac{79.546gCuO}{1molCuO} =0.078gCuO$

Therefore, the percent yield is:

$Y=\frac{0.038g}{0.078g}*100\%\\ \\Y=48.6\%$

Best regards.

6 0

3 years ago