You should clean up after every investigation because if you leave a mess, maybe another detective will come in and get lost because of the mess you left.
Answer:
1. Equivalence point
2. Direct titration
3. Primary standard
4. Titrand
5. Back titration
6. Standard solution
7. Titrant
8. Indirect titration
9. End point
10. Indicator
Explanation:
1. The equivalence point is the tiration point at which the quantity or moles of the added titrant is sufficient or equal to the quantity or moles of the analyte for the neutralization of the solution of the analyte.
2. Direct titration is a method of quantitatively determining the contents of a substance
3. A primary standard is an easily weigh-able representative of the mount of moles contained in a substance
4. A titrand is the substance of unknown concentration which is to be determined
5. The titration method that uses a given amount of an excess reagent to determine the concentration of an analyte is known as back titration
6. A standard solution is a solution of accurately known concentration
7. A titrant is a solution that has a known concentration and which is titrated unto another solution to determine the concentration of the second solution
8. Indirect titration is the process of performing a titration in athe reverse order
9. The end point is the point at which the indicator indicates that the equivalent quantities of the reagents required for a complete reaction has been added
10 An indicator is a compound used to visually determine the pH of a solution.
Physical isn't so chemical
If a sample of gas is a 0.622-gram, volume of 2.4 L at 287 K and 0.850 atm. Then the molar mass of the gas is 7.18 g/mol
<h3>What is an ideal gas equation?</h3>
The ideal gas law (PV = nRT) relates to the macroscopic properties of ideal gases.
An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).
Given :
The ideal gas equation is given below.
n = PV/RT
n = 86126.25 x 0.0024 / 8.314 x 287
n = 0.622 / molar mass (n = Avogardos number)
Molar mass = 7.18 g
Hence, the molar mass of a 0.622-gram sample of gas having a volume of 2.4 L at 287 K and 0.850 atm is 7.18 g
More about the ideal gas equation link is given below.
brainly.com/question/4147359
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It takes 21.3 days
<h3>Further explanation</h3>
Given
5 hr = 8 kg Alcohol
Required
Days to consume 1000 kg of glucose
Solution
Alcoholic fermentation⇒ glucose produces ethanol and carbon dioxide,
C₆H₁₂O₆ → 2 C₂H₅OH + 2CO₂
mol ethanol :
moles of glucose to produce 108.7 moles ethanol :
54.35 moles = 5 hours
moles of 1000 kg of glucose :
So for 5555.5 moles, it takes :
