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Setler79 [48]
2 years ago
7

Explain why “advancing water contact angles are more sensitive to the hydrophobic property of a surface whereas receding water c

ontact angles are more sensitive to the hydrophilic property of a surface”.
Chemistry
1 answer:
timama [110]2 years ago
7 0

Answer: Surface tension has a more complicated effect on capillary pressure: on the one hand, capillary pressure is directly proportional to surface tension. High surface tension liquids, on the other hand, typically have higher contact angles, which lowers capillary pressure.

Explanation: Does this help??

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Which of the following is/are correct regarding the pentose phosphate pathway? I. NADH is generated by the oxidation of glucose-
fgiga [73]

Answer:

The correct answer is 1 NADH is generated by the oxidation of glucose-6-phosphate.

Explanation:

Pentose phosphate pathway deals with the utilization of glucose-6-phosphate by oxidation process to form 6-phosphogluconolactone by the catalytic activity of glucose-6-phosphate dehydrogenese.

   This enzyme need NAD+ as co enzyme which get reduced to generate NADH.

4 0
3 years ago
Carbohydrates are formed by plants converting water a d carbon dioxide into glucose and oxygen, in the photocatalyzed process is
Dmitry [639]
Photosynthesis maybe.
3 0
2 years ago
What is the origin of the name for bismuth
Juliette [100K]
The name come from the German 'Bisemutum' a corruption of 'Weisse Masse' meaning white mass. 
6 0
3 years ago
A compound is found to contain 73.23% xenon name 26.77% oxygen by mass. What is the empirical formula for this compound ?
Luba_88 [7]

The empirical formula is XeO₃.

<u>Explanation:</u>

Assume 100 g of the compound is present. This changes the percents to grams:

Given mass in g:

Xenon = 73.23 g

Oxygen = 26.77 g

We have to convert it to moles.

Xe = 73.23/   131.293 = 0.56 moles

O = 26.77/ 16 = 1.67 moles

Divide by the lowest value, seeking the smallest whole-number ratio:

Xe = 0.56/ 0.56 = 1

O = 1.67/ 0.56 = 2.9 ≈3

So the empirical formula is XeO₃.

6 0
3 years ago
How much salt (NaCl) is carried by a river flowing at 30.0 m3/s and containing 50.0 mg/L of salt
denis23 [38]

Answer:

129,600kg/day

Explanation:

The river is flowing at 30.0

1 m³ is equivalent to 1000L

flowrate of river = 30*1000 =30,000L/s

Convert L/s to litre per day by multiplying by 24*60*60

flowrate of river = 30,000 * 24*60*60 L/day

                    = 2,592,000,000L/day

if the river contains 50mg of salt  in 1L of solution

lets find how many mg of salt Y is contained in 2,592,000,000L/day

by cross multiplying we have

Y=\frac{2592000000*50}{1}

Y= 129,600,000,000 mg/day

convert this value to kg/day by dividing by 1 million

Y= 129,600,000,000/1000000

Y= 129,600kg/day

3 0
2 years ago
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