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kenny6666 [7]
3 years ago
10

A gas has a volume of 1.4 liters at STP. What is the new volume when the temperature is changed to 40 K and the pressure is chan

ged to 1.2 atm?
Chemistry
1 answer:
Whitepunk [10]3 years ago
7 0

Answer:

0.17094 liters

Explanation:

-For STP, Pressure=1 atm and Temperature=273K

-We apply the combined gas law which relates temperature, pressure and Volume:

\frac{PV}{T}=k

-We therefore equate the relationship at different temperatures as follows:

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\\\\\frac{1\times 1.4\ }{273}=\frac{1.2\times V_2}{40}\\\\V_2=\frac{1\times 1.4\ }{273}\times \frac{40}{1.2}\\\\=0.17094\ L

Hence, the new volume is 0.17094 liters

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What are cells at express the same gene
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Hi There!

Question - What are cells at express the same gene?

Answer - A. the same

Why - Every single cell in a person body has the exact same DNA.

Hope This Helps :)

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3 years ago
What element is most likely to form a diatomic molecule?
Harman [31]
Bromine (Br) is the only element listed that is a diatomic element. I believe that Bromine is going to be your answer.
5 0
4 years ago
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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
Oliga [24]

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.

3 0
3 years ago
What is the pH of a 6.7 x 10^−5 M H+ solution?
lyudmila [28]

Answer:

pH = 4.17

Explanation:

According to the molar concentration you stated, pH of the solution is: 4.17

Remember that pH = - log [H⁺]

and [H⁺] = 10^-pH

When:

pH > 7 →  Basic solution

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pH < 7 → Acid solution

3 0
4 years ago
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Realice las siguientes conversiones: a) 72°F a °C, b) 213.8°C a °F, c)180°C a K, d) 315K a °F, e) 1750°F a K, f) 0K a °F.
bonufazy [111]

Answer:

a) 72 °F= 22.22 °C

b)  213.8 °C=  416.84°F

c) 180 °C= 453.15 °K

d) 315 °K=  107.33 °F

e) 1750 °F= 1227.594 °K

f) 0 °K=  -459.67 °F

Explanation:

Para realizar el intercambio de unidades debes tener en cuenta las siguientes conversiones:

  • Fahrenheit a Celsius: C=\frac{F-32}{1.8}
  • Celsius a Fahrenheit: °F= °C*1.8 + 32
  • Celsius a Kelvin: °K= °C + 273.15
  • Kelvin a Fahrenheit: F= (K -273.15)*1.8 + 32
  • Fahrenheit a Kelvin:K=\frac{F-32}{1.8} + 273.15

Entonces se obtiene:

a) 72 °F= \frac{72-32}{1.8}=22.22 °C

b)  213.8 °C= 213.8*1.8 + 32= 416.84°F

c) 180 °C= 180°C + 273.15= 453.15 °K

d) 315 °K= (315 -273.15)*1.8 + 32= 107.33 °F

e) 1750 °F= \frac{1750-32}{1.8} + 273.15= 1227.594 °K

f) 0 °K= (0 -273.15)*1.8 + 32= -459.67 °F

7 0
3 years ago
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