Answer:
The answer to your question is 25.2 g of acetic acid.
Explanation:
Data
[Acetic acid] = 0.839 M
Volume = 0.5 L
Molecular weight = 60.05 g/mol
Process
1.- Calculate the number of moles of acetic acid
Molarity = moles / volume
-Solve for moles
moles = Molarity x volume
-Substitution
moles = (0.839)(0.5)
-Result
moles = 0.4195
2.- Calculate the mass of acetic acid using proportions and cross multiplications
60.05 g ----------------------- 1 mol
x ----------------------- 0.4195 moles
x = (0.4195 x 60.05) / 1
x = 25.19 g
3.- Conclusion
25.2 g are needed to prepare 0.500 L of Acetic acid 0.839M
Answer:
0.97 kg, 1.0kg
Explanation:
As you have mentioned in the question, to convert kilo-gram to grams, this conversion is used, 1 kilogram = 1,000 grams
Initially, when one has to convert from one unit to another, there is a lot of confusion. To avoid any such, multiply and divide the number with units to which it has to be converted.
Here,
970 g = 
This makes no difference to the number.
970 g = 
As, 1 kg= 1,000 g
970 g = 
970 g = 0.97 kg
Rounding this off to the nearest tenth of a kilogram.
The tenths place is immediately to the right of the decimal point.
Here, 9 is at the tenth's place. Round the tenth's place up if the digit at hundredth's place is greater than or equal to 5. Since, 7 > 5, up the number at tenth's place.
So, answer is 1.0 kg
Answer:: We're asked to find the molar concentration of the NaCl solution given some titration data.
Explanation:
Answer:
50%
Explanation:
Hydrogen makes up about 50% in glucose in atoms, in mass, 6.7 wise
Thin layer chromatography(TLC) works with the principle of separation through adsorption.
It is used in the isolation and extraction of lipids through the following steps:
- apply the lipid samples spots in the bottom of the plate.
- also apply sample solution to the marked spot
- pour the mobile phase into the TLC chamber and use a moist filter paper to cover it. this is done to maintain equal humidity.
- then place the plate in the TLC chamber and close it with a lid.
- the plate is immersed into the solvent (mobile phase) for its development. this is done, keeping in mind that the sample spot should be above the solvent.
- once the sample spots are developed, they are removed and dried.
- this is later viewed using the UV light chamber to see the isolation of the lipid sample.
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brainly.com/question/3137660