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Ahat [919]
2 years ago
11

Rate how successful you were at using Active Reading skills by responding to the following statement: As i read, i paid attentio

n to my own understanding and recorded my thoughts and questions.​
Chemistry
1 answer:
Aleksandr-060686 [28]2 years ago
8 0

It is possible to rate your active reading skills either by using a number is a specific rate or by using a word that qualifies your performance.

<h3>What are active reading skills?</h3>

Active reading skills are a set of skills that help you read actively; this means skills to effectively read a text and understand it.

<h3>How to rate your reading skills?</h3>

In this case, you are asked to rate your ability to pay attention to your understanding, thoughts, and questions while reading. To do this you can:

Rate using numbers:

  1. Select a specific scale rating such as 1 to 5 or 1 10.
  2. Analyze honestly if you used this ability while reading the text.
  3. Give yourself a number. For example, if you apply this skill only sometimes while reading you can rate yourself 5 out of 10.

Rate using words:

  1. Select the types of words that you will use, the most common ones are frequency words such as always, sometimes, and never.
  2. Analyze honestly your performance.
  3. Find the word that best fits it.

Learn more about rate in: brainly.com/question/2278897

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Explanation:

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A 0.944 M solution of glucose, C6H12O6, in water has a density of 1.0624 g/mL at 20∘C. What is the concentration of this solutio
____ [38]

Answer:

Mole fraction: 0,0157.

Molality: 0,889m

Mass%: 16%

Explanation:

<em>The units are mole fraction, molality and mass%</em>

<em />

Units of mole fraction are moles of glucose per total moles.

Moles of glucose assuming 1L are:

0,944 moles.

Moles of water in 1L are:

1L × (1,0624kg/L) × (1000g / 1kg) × (1mol / 18,02g) = <em>59,0 moles of water</em>

<em />

Mole fraction is: 0,944 moles / (59,0 mol + 0,944mol) = <em>0,0157</em>

Molality is mole of solute (0,944) per kg of solution (1,0624kg):

0,944mol / 1,0624kg = <em>0,889m</em>

<em></em>

In mass percent total mass is 1062,4g and mass of 0,944 moles of glucose is:

0,944mol×(180,156g/1mol) = 170g of glucose. Mass%:

170g / 1062,4g ×100 = <em>16%</em>

<em></em>

I hope it helps!

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