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SashulF [63]
3 years ago
13

10. When a 13.6 g sample of a compound containing only magnesium and oxygen is decomposed, 5.4 g of oxygen is obtained. What is

the percent of
magnesium in this sample?
• 5.4 g / 13.6 g * 100
• (13.6 g- 5.4g) / 13.6 g x 100
• (13.6 g- 5.4g) x 100
• 5.4g/ (13.6 g-5.4g) x 100
Chemistry
2 answers:
kari74 [83]3 years ago
8 0

Answer: • (13.6 g- 5.4g) / 13.6 g x 100 is the correct option

Explanation:

Total mass of the sample containing only magnesium and oxygen is 13.6 grams. This means that the sum of the mass of oxygen and the mass of magnesium is 13.6 grams.

On decomposition of the sample, 5.4 g of oxygen is obtained. This means that the mass of magnesium that was obtained would be

13.6 - 5.4 = 8.2 grams of magnesium. Therefore, the percentage of magnesium in this sample would be

8.2/13.6 × 100 = 60.294%

irinina [24]3 years ago
7 0

Answer:5.4 g / 13.6 g *100

Explanation:Its is the correct answer

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irvinase is an enzyme that has 4 cys residues tied up in 2 disulfide bonds. you denature irvinase with 8m urea in the presence o
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Answer:

1. Quaternary structure of proteins relates to the interactions between separate polypeptide chains within the protein. The word polypeptide refers to a polymer of amino acids. A protein may contain one or more polypeptides and is folded and may be covalently modified.

2. Hemoglobin (and many other proteins) have multiple polypeptide subunits. Interactions between the subunits include ionic interactions, hydrogen bonds, and hydrophobic interactions. Modification of the quaternary structure of a protein may have the same effects as modification of its tertiary structure - alteration of its function/activity.

3. The enzyme ribonuclease (RNase) is interesting in being very stable to heat and other things that denature/inactivate other proteins. (By the way, denaturation is a word that means the tertiary and/or quaternary structure of a protein is disrupted.). RNase has disulfide bonds that help it to remain resistant to denaturation. Heating it to 100 Celsius, which denatures most proteins does not denature RNase. Breaking the disulfide bonds of RNAse with a reagent like mercaptoethanol followed by heating to 100 Celsius to destroy hydrogen bonds (or treatment with urea) causes loss of activity. If one allows the hydrogen bonds to reform slowly, some of the enzyme's activity reappears, which indicates that the information necessary for proper folding is contained in the primary structure (amino acid sequence).

4. Disulfide bonds are important structural components of proteins. They form when the sulfhydryls of two cysteines are brought together in close proximity. Some chemicals, such as mercaptoethanol, can reduce the disulfides (between cysteine residues) in proteins to sulfhydryls. In the process of transferring electrons to the cysteines, the sulfhydryls of mercaptoethanol become converted to disulfides. Treatment of RNase with mercaptoethanol reduces RNAse's disulfides to sulfhydryls. Subsequent treatment of RNase with urea disrupts hydrogen bonds and allows the protein to be denatured.

5. Interestingly, removal of the mercaptoethanol and urea from the solution allows RNase to refold, reestablish the correct disulfide bonds, and regain activity. Clearly, the primary sequence of this protein is sufficient for it to be able to refold itself to the proper configuration.

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8. Proteins sometimes have amino acids in them that are chemically modified. Chemical modification of amino acids in proteins almost always occurs AFTER the protein is synthesized (also described as post-translational modification). Examples include hydroxyproline and hydroxylysine in collagen, gamma carboxyglutamate, and phosphoserine. Modification of the collagen residues allows for the triple helical structure of the protein and for the strands to be cross-linked (an important structural consideration).

9. Hemoglobin (and many other proteins) have multiple polypeptide subunits. Interactions between the subunits include disulfide bonds, ionic interactions, hydrogen bonds, hydrophilic, and hydrophobic interactions. Modification of the quaternary structure of a protein may have the same effects as modification of its tertiary structure - alteration of its function/activity.

10. Folding is necessary for proteins to assume their proper shape and function. The instructions for folding are all contained in the sequence of amino acids, but we do not yet understand how those instructions are carried out rapidly and efficiently. Levinthal's paradox illustrates the fact that folding is not a random event, but rather based on an ordered sequence of events arising from the chemistry of each group.

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8 0
3 years ago
For the reaction, identify the Lewis acid and the Lewis base.<br> AlF3 + CH3F → CH3+ + [AlF4]-
Ivanshal [37]
Answer:
            AlF₃ is Lewis Acid

            CH₃F is Lewis Base

Explanation:
                     According to Lewis concept ,"those compounds which donate pair of electrons are called as Lewis Base and those accepting pair of electrons are called as Lewis Acid.

In Given Reaction,
<span>                               AlF</span>₃<span> + CH</span>₃<span>F → CH</span>₃⁺<span> + [AlF</span>₄<span>]</span>⁻

AlF₃ is acting as acid because the octet of Al is not complete, hence it has tendency to accept electrons.

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5 0
2 years ago
One mL of water is equal to
solniwko [45]

1 milliliter of water (ml)

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Fraction : 1/25 ounces of water (oz wt.)

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3 years ago
Why are iron needles used in compasses
ludmilkaskok [199]

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4 0
2 years ago
How much heat is required to heat 1.6g of ice from -16c to steam at 112c?
oksian1 [2.3K]

Answer:

Total heat ≅ 49.07 kJ

Explanation:

Given that:

mass = 1.6 g = 0.016 kg

Initial temperature = - 16 ° C

final temperature = 112° C

specific heat for ice = 2.06 kJ/kgC

specific heat of water = 4.186 kJ/kgC

heat fusion of ice = 334 kJ/kg

specific heat for steam = 2.1 kJ/kgK

heat of vaporization of water = 2256 kJ/kg

To heat ice from -16 ° C to 0 ° C

Q₁ =  2.06 kJ/kgC × 0.016 kg ×  16 ° C

Q₁ =  0.52736 kJ

To melt Ice at 0° C

Q₂= 334 kJ/kg × 0.016 kg = 5.344 kJ

To heat water from 0° C to  100° C

Q₃ = 4.186 kJ/kgC × 0.016 kg  × 100° C

Q₃ = 6.6976 kJ

To vaporize water to steam at 100° C

Q₄ = 2256 kJ/kg × 0.016 kg = 36.096 kJ

To heat steam from 100C to 112° C

Q₅ = 2.1 kJ/kgC × 0.016 kg × 12 C

Q₅ = 0.4032 kJ

Total heat = Q₁ + Q₂ + Q₃ + Q₄  + Q₅

Total heat =  (0.52736 +  5.344 +  6.6976 + 36.096 + 0.4032) kJ

Total heat = 49.06816  kJ

Total heat ≅ 49.07 kJ

6 0
3 years ago
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