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weqwewe [10]
2 years ago
12

What would you multiply "moles of oxygen" by to get the units "grams of oxygen"?

Chemistry
1 answer:
photoshop1234 [79]2 years ago
6 0

Answer:

AM

Explanation:

to go from moles to grams you multiply by the Atomic Mass or Molar Mass (Atomic Mass for an element and Molar Mass for a compound).

The formula is:

Mass = moles * MM

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heat and pressure

Explanation:

A metamorphic rock is a changed rock. It is formed a result of mineral changes in a pre-existing rock in the presence of temperature and pressure conditions.

Metamorphic transformation in rocks begins to occur when minerals starts to change from their original and usual form. Most metamorphic rocks are formed as a result of burial into deep seated environment or when subjected to conditions that can elevate the temperature and pressure in a rock.

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2 years ago
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Which of the following is considered a recessive trait?
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What is the mass of a sample of NH3 containing 7.20 x 1024 molecules of NH3? (5 points) Group of answer choices 203 grams 161 gr
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Answer:

203 grams

Explanation:

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The no. of grams of NH₃ present = no. of moles x molar mass = (11.96 mol)(17.0 g/mol) = 203.3 g ≅ 203.0 g.

8 0
3 years ago
Hydrogen bonding between polyamide chains plays an important role in determining the properties of a nylon such as nylon 6,6. Dr
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Answer:

See figure 1

Explanation:

In the structure of nylon 6,6 we have <u>amide groups</u>. In this functional group, We have a nitrogen bond to hydrogen, so in this bond, we will have a <u>dipole</u>, due to the <u>electronegativity difference</u>. Nitrogen has more electronegativity than hydrogen, therefore a <u>positive dipole</u> would be generated in the hydrogen atom. Additionally, in the <u>carbonyl group</u> (C=O) due to the oxygen, we will have also a <u>dipole</u>, in this case, a <u>negative dipole</u> because the oxygen atom has <u>more electronegativity</u> (compare with carbon).

When we put two strings of nylon 6,6 the positive dipole will interact with the negative dipole and vice-versa and we will obtain the <u>"hydrogen bonds"</u>.

See figure 1

I hope it helps!

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