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Nataly_w [17]
3 years ago
14

2mL of a serum sample was added to 18mL of phosphate buffered saline (PBS) in Tube 1. 10mL of Tube 1 was added to 40mL of PBS in

Tube What is the dilution of serum in Tube 2?
Chemistry
1 answer:
Sladkaya [172]3 years ago
4 0

Answer:

Tube 2 has a total dilution of 1:50

Explanation:

We have a 2 ml serum sample added to a 18 mL phosphate buffered saline sample in tube 1. This means now in tube 1 there is 20 mL.

We have a 1:10 (= 2:20) dilution here.

10 ml of this 1:10 diluted tube 1 is taken and added to a 40 mL of PBS in tube 2.

Now we have 50 mL in tube 2.

This is a 10:50 (= 1:5) dilution.

The total dilution is 10x5 = 50

So the total ditultion has a rate 1:50

Tube 2 has a total dilution of 1:50

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Alborosie

Answer:

n=0.430molH_2

Explanation:

Hello!

In this case, considering the partial Dalton's law of partial pressures, we can notice that the total pressure equals the pressure of steam and the pressure of hydrogen, which can be determined as shown below:

p_T=p_H+p_w\\\\p_H=811torr-12torr=799torr*\frac{1atm}{760torr}\\\\p_H=1.05atm

Thus, by using the ideal gas law, we can compute the moles of hydrogen as shown below:

PV=nRT\\\\n= \frac{PV}{RT}=\frac{1.05atm*10.0L}{0.082\frac{atm*L}{mol*K}*298K}\\\\n=0.430molH_2

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3 years ago
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The volume of 10 grams of frozen water is more than the volume of 10 grams of liquid water
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What is the energy of a photon of red light that has a frequency of 4.48 X 10^14Hz?
lana [24]

Answer:

E = 29.7× 10⁻²⁰  j

Explanation:

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h = planck's constant

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E = 6.63 × 10⁻³⁴ Kg.m² /s × 4.48 × 10¹⁴ s⁻¹

E = 29.7× 10⁻²⁰ Kg.m²/s²

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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:

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<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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2 years ago
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