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inessss [21]
2 years ago
10

if you were conducting an experiment with pepsin which has an optimal enzymatic actigity at ph 2.3, wat buffer would be the best

choice
Chemistry
2 answers:
ra1l [238]2 years ago
6 0

Answer:

Explanation:   Is there choises? H3PO4 / NaH2PO4 buffer with equal concentrations would form a buffer solution pH = 2.15

forsale [732]2 years ago
3 0

Answer: One with a pKa of 1.9

Hope this helps <3

P.S Fun Fact~~

There are only two words in the English language that have all five vowels in order: "abstemious" and "facetious."!

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I need help with chemical formulas<br>​
pochemuha

Answer:

Which one?

Explanation:

7 0
2 years ago
Recessive genes are _____.
AveGali [126]

Answer:

1) represented by a lower case letter
2) always masked by dominant genes

3) only expressed in the absence of a dominant gene

Explanation:

So if I take a black and a yellow marker- and tried drawing the yellow over the black- or the other way around- you'd still only see the black marker. The recessive trait is like the yellow marker. Now if I drew the yellow marker with another yellow marker you would see the yellow. I hope this helps!

8 0
2 years ago
Which choice best explains how herd behavior helps young musk oxen survive?
laiz [17]

Answer:

A

Explanation:

'herd behavior' helps understand the question for the answer

7 0
2 years ago
Read 2 more answers
A mixture of 75 mole% methane and 25 mole% hydrogen is burned with 25% excess air. Fractional conversions of 90% of the methane
son4ous [18]

Solution :

Consider a mixture of methane and hydrogen.

Take the basis as 100 moles of the mixture.

The mixture contains 75% of methane and 25% of hydrogen by mole and it is burned with 25% in excess air.

Moles of methane = 0.75 x 100

Moles of hydrogen = 0.25 x 100

The chemical reactions involved during the reaction are :

$CH_4+2O_2 \rightarrow CO_2 + 2H_2O$

$CH_4+1.5O_2 \rightarrow CO+2H_2O$

$H_2+0.5O_2 \rightarrow H_2O$

The fractional conversion of methane is 90%

Number of moles of methane burned during the reaction is = 0.9 x 75

                                                                                                   = 67.5

Moles of methane leaving = initial moles of methane - moles of methane burned

                                           = 75 - 67.5

                                           = 7.5 moles

Fractional conversion of hydrogen is 85%

The number of moles of hydrogen burned during the reaction is = 0.85 x 25

                                                                                                   = 21.25

Moles of hydrogen leaving = initial moles of hydrogen - moles of hydrogen burned

                                           = 25 - 21.25

                                           = 3.75 moles

Methane undergoing complete combustion is 95%.

$CO_2$ formed is = 0.95 x 67.5

                       = 64.125 moles

$CO$ formed is = 0.05 x 67.5

                       = 3.375 moles

Oxygen required for the reaction is as follows :

From reaction 1, 1 mole of the methane requires 2 moles of oxygen for the complete combustion.

Hence, oxygen required is = 2 x 75

                                            = 150 moles

From reaction 3, 1 mole of the hydrogen requires 0.5 moles of oxygen for the complete combustion.

Hence, oxygen required is = 0.5 x 25

                                            = 12.5 moles

Therefore, total oxygen is = 150 + 12.5 = 162.5 moles

Air is 25% excess.

SO, total oxygen supply = 162.5 x 1.25 = 203.125 moles

Amount of nitrogen = $203.125 \times \frac{0.79}{0.21} $

                                = 764.136 moles

Total oxygen consumed = oxygen consumed in reaction 1 + oxygen consumed in reaction 2 + oxygen consumed in reaction 3

Oxygen consumed in reaction 1 :

1 mole of methane requires 2 moles of oxygen for complete combustion

 = 2 x 64.125

 = 128.25 moles

1 mole of methane requires 1.5 moles of oxygen for partial combustion

= 1.5 x 3.375

= 5.0625 moles

From reaction 3, 1 mole of hydrogen requires 0.5 moles of oxygen

= 0.5 x 21.25

= 10.625 moles.

Total oxygen consumed = 128.25 + 5.0625 + 10.625

                                        = 143.9375 moles

Total amount of steam = amount of steam in reaction 1 + amount of steam in reaction 2 + amount of steam in reaction 3

Amount of steam in reaction 1 = 2 x 64.125 = 128.25 moles

Amount of steam in reaction 2 = 2 x 3.375 = 6.75 moles

Amount of steam in reaction 3  = 21.25 moles

Total amount of steam = 128.25 + 6.75 + 21.25

                                     = 156.25 moles

The composition of stack gases are as follows :

Number of moles of carbon dioxide = 64.125 moles

Number of moles of carbon dioxide = 3.375 moles

Number of moles of methane = 7.5 moles

Number of moles of steam = 156.25 moles

Number of moles of nitrogen = 764.136 moles

Number of moles of unused oxygen = 59.1875 moles

Number of moles of unused hydrogen = 3.75 moles

Total number of moles of stack  gas

= 64.125+3.375+7.5+156.25+764.136+59.1875+3.75

= 1058.32 moles

Concentration of carbon monoxide in the stack gases is

$=\frac{3.375}{1058.32} \times 10^6$

= 3189 ppm

b).  The amount of carbon monoxide in the stack gas can be decreased by increasing the amount of the excess air. As the amount of the excess air increases, the amount of the unused oxygen and nitrogen in the stack gases will increase and the concentration of CO will decrease in the stack gas.  

6 0
3 years ago
The chemical bonds of carbohydrates and lipids have high potential energy because:
gladu [14]

Answer:

c. Many of their bonds are C-C and C-H

Explanation:

The majority of bonds in  carbohydrates and lipids( being an organic compound) are C-C and C-H. Like glucose, fructose or galactose ,etc.

These bonds are strong and do require a lot of energy to break. Thus, a lot of energy are required to break carbs and lipids into simpler compounds.Therefore, carbohydrates and lipids have high potential energy.

The correct answer is c.

4 0
2 years ago
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