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vovangra [49]
3 years ago
9

The diagram show the box for an element in the periodic table . What is the atomic mass of the element shown .

Chemistry
1 answer:
DochEvi [55]3 years ago
8 0

Answer:32.1

Explanation:

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If the same amount of heat is added to 25.0 g of each of the metals, which are all at the same initial temperature, which metal
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Answer:

The bismuth sample.

Explanation:

The specific heat c of a substance (might not be a metal) is the amount of heat required for heating a unit mass of this substance by unit temperature (e.g., \rm 1\; ^{\circ}C.) The formula for specific heat is:

\displaystyle c = \frac{Q}{m \cdot \Delta T},

where

  • Q is the amount of heat supplied.
  • m is the mass of the sample.
  • \Delta T is the increase in temperature.

In this question, the value of Q (amount of heat supplied to the metal) and m (mass of the metal sample) are the same for all four metals. To find \Delta T (change in temperature,) rearrange the equation:

\displaystyle c \cdot \Delta T = \frac{Q}{m},

\displaystyle \Delta T = \frac{Q}{c \cdot m}.

In other words, the change in temperature of the sample, \Delta T can be expressed as a fraction. Additionally, the specific heat of sample, c, is in the denominator of that fraction. Hence, the value of the fraction would be the largest for sample with the smallest specific heat.

Make sure that all the specific heat values are in the same unit. Find the one with the smallest specific heat: bismuth (\rm 0.123 \; J \cdot g\cdot \,^{\circ}C^{-1}.) That sample would have the greatest increase in temperature. Since all six samples started at the same temperature, the bismuth sample would also have the highest final temperature.

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3 years ago
In glycolysis, if glucose is labeled at the carbon 6 position (see page 1 for numbering of carbons in glucose) A) the carbon wit
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Answer:

D) the carbon with the low-energy phosphate on it in 1,3 BPG is labeled.

Explanation:

Glycolysis has 2 phase (1) preparatory phase (2) pay-off phase.

<u>(1) Preparatory phase</u>

During preparatory phase glucose is converted into fructose-1,6-bisphosphate. Till this time the carbon numbering remains the same i.e. if we will label carbon at 6th position of glucose, its position will remian the same in fructose-1,6-bisphosphate that means the labeled carbon will still remain at 6th position.

When fructose-1,6-bisphosphate is further catalyzed with the help of enzyme aldolase it is cleaved into two 3 carbon intermediates which are glyceraldehyde 3-phosphate (GAP) and dihyroxyacetone  phosphate (DHAP).  In this conversion, the first three carbons of fructose-1,6-bisphosphate become carbons of DHAP while the last three carbons of fructose-1,6-bisphosphate will become carbons of GAP. It simply means that GAP will acquire the last carbon of fructose-1,6-bisphosphate which is labeled. Now the last carbon of GAP which has phosphate will be labeled.  

<u>(2) Pay-off phase</u>

During this phase, GAP is dehydrogenated into 1,3-bisphosphoglycerate (BPG) with the help of enzyme glyceraldehyde 3-phosphate dehydrogenase. This oxidation is coupled to phosphorylation of C1 of GAP and this is the reason why 1,3-bisphosphoglycerate has phosphates at 2 positions i.e. at position 1 in which phosphate is newly added and position 3rd which already had labeled carbon.

It is pertinent to mention here that<u> BPG has a mixed anhydride and the bond at C1 is a very high energy bond.</u> In the next step, this high energy bond is hydrolyzed into a carboxylic acid with the help of enzyme phosphoglycerate kinase and the final product is 3-phosphoglycerate. Hence, the carbon with low energy phosphate i.e. the carbon at 3rd position remains labeled.

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Explanation:

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