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fenix001 [56]
3 years ago
14

how would it affect your results if you use a wet Erlenmeyer flask instead of a dry one when transferring your acid solution fro

m a volumetric pipette? pls someone answer pls​
Chemistry
2 answers:
zloy xaker [14]3 years ago
8 0
With a wet Erlenmeyer flask, the result will not be accurate. It might be contaminated by any type of liquid such as basic or acidic. Therefore, an error will occur since it is not 100% acid.- hope this helped
solniwko [45]3 years ago
8 0

Explanation:

Your results would be effected because there is a possibility that the water and chemical substance added into the flask will combine, possibly creating a reaction, either making the solution stronger or weaker, or possibly just messing up your experiment in one go.

For example:

If I paint a picture, and I have a cup that I use in order to put paint in, but the cup is wet and I put my paint in it, my paint will become more liquefied due to the fact water has been added to it, messing the paint up.

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Alexandra decides to climb Mt. Krumpett, which is 5000 m high. She determines that this will require a total of 1350 kcal of ene
Kisachek [45]
Alexandra requires a total energy of 1350 kcal for the climb
by eating proteins, fats and carbohydrates the amount of calories per gram contributed varies.
Proteins and carbohydrates - 4 calories per gram 
fats - 9 calories and gram
This means that by eating the same mass of fats and proteins/ carbohydrats the calories gained from fats is higher.
each bar contains;
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40 g of protein - 4 calories/g x 40 g = 160 calories 
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8 0
3 years ago
Which of the following is not a part of every scientific investigation?
Dmitry_Shevchenko [17]

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6 0
3 years ago
Carbon tetrachloride, CCl4, was once used as a dry cleaning solvent, but is no longer used because it is carcinogenic. At 57.8 °
Mila [183]
This problem is to use the Claussius-Clapeyron Equation, which is:

ln [p2 / p1] = ΔH/R [1/T2 - 1/T1]

Where p2 and p1 and vapor pressure at estates 2 and 1

ΔH is the enthalpy of vaporization

R is the universal constant of gases = 8.314 J / mol*K

T2 and T1 are the temperatures at the estates 2 and 1.

The  normal boiling point => 1 atm (the pressure of the atmosphere at sea level) = 101,325 kPa

Then p2 = 101.325 kPa
T2 = ?
p1 = 54.0 kPa
T1 = 57.8 °C + 273.15K = 330.95 K
ΔH = 33.05 kJ/mol = 33,050 J/mol 

=> ln [101.325/54.0] = [ (33,050 J/mol) / (8.314 J/mol*K) ] * [1/x - 1/330.95]

=> 0.629349 = 3975.22 [1/x - 1/330.95] = > 1/x =  0.000157 + 1/330.95 = 0.003179

=> x = 314.6 K => 314.6 - 273.15 = 41.5°C

Answer: 41.5 °C 
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3 years ago
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