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2 years ago

What type of large molecule is responsible for repair and growth of cells?

Chemistry

2 answers:

goldfiish [28.3K]2 years ago

8 0

Answer:

$\boxed {\boxed {\sf A.\ Protein}}$

Explanation:

Macromolecules play many important roles in the body. Let's list the functions of proteins, carbohydrates, and lipids.

Proteins:

These molecules help with growth, building and repairing cells, transporting substances, and acting as enzymes or hormones.

Carbohydrates:

These molecules are used for energy. They can provide quick energy, because they are easy to break down.

Lipids:

These help store energy for later use, plus they are sometimes insulators and hormones.

Based on the information, the molecule responsible for repair and growth must be <u>protein.</u>

Oliga [24]2 years ago

7 0

Answer:

The answer is a. proteins

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How many grams of water can be produce from the complete reaction of excess nitric acid and 33.2 mL of 0.245 M lithium hydroxide

Mariulka [41]

Mass of water produced : 0.146 g

<h3>Further explanation</h3>

Given

33.2 mL of 0.245 M lithium hydroxide

Required

mass of water

Solution

Reaction

HNO₃ (aq) + LiOH (aq) → H₂O (l) + LiNO₃ (aq)

mol LiOH :

= M x V

= 0.245 x 33.2 ml

= 8.134 mmol

From the equation, the mol ratio of HNO₃ : H₂O = 1 : 1, so mol H₂O = 8.134 mmol

mass H₂O :

= mol x MW

= 8.134 x 10⁻³ mol x 18 g/mol

= 0.146 g

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2 years ago

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Explanation:

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2 years ago

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? Answer the question below. Type your response in the space provided. What volume of a 2.5 M stock solution of acetic acid (HC2

Novay_Z [31]

Data:
$M_{concentrated} = 2.5\:mol$
$V_{concentrated} = ?$
$M_{dilute} = 0.50\:mol$
$V_{dilute} = 100\:mL\to0.100\:L$
<span>
Formula: Dilution Calculations

</span> $M_{concentrated} * V_{concentrated} = M_{dilute} * V_{dilute}$
<span>
Solving:

</span>
$M_{concentrated} * V_{concentrated} = M_{dilute} * V_{dilute}$
$2.5 * V_{concentrated} = 0.50 * 0.100$
$2.5V_{concentrated} = 0.05$
$V_{concentrated} = \frac{0.05}{2.5}$
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3 years ago

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ipn [44]

Answer:

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2 years ago

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3. A 4.00 gram sample of solid gold was heated from 274K to 314K. How much energy was involved?

MrMuchimi

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2 years ago