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aleksklad [387]
3 years ago
9

In what situation do we use a volumetric flask, conical flask, pipette and graduated cylinder? Explain your answer from accuracy

aspects of these apparatus.
Chemistry
1 answer:
Ad libitum [116K]3 years ago
3 0
A volumetric flask is used to contain a predetermined volume of substance and only measures that volume, for example 250 ml.
Conical flasks can be used to measure the volume of substances but the accuracy they provide is usually up to 10ml. Conical flasks are used in titrations, reactions where the liquid may boil, and reactions which involve stirring. 
Pippettes are of two types, volumetric and graduated. Pippettes are used where high accuracy is required and volumetric pippettes come in as little as 1 ml. Pippettes are usually used in titrations.
Graduated cylinders come in a wide variety of sizes and their accuracy can be down to as much as 1 ml. They are used to contain liquids.
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Nostrana [21]
3) +7

I got this stupid question wrong so that you people don't have to. I simply hate chemistry and I wish it didn't exist. Kind of like this website that makes me explain the answer. 

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4 0
3 years ago
consider multiplying 26.2 by 16.43. what would a mathematician say the answer is? what would a scientist say the answer is?
levacccp [35]

Answer:

Well they both would say numbers

Explanation:

The reason why is they both speak english

4 0
3 years ago
What mass of Ba3(PO4)2 is contained in 1575 mL of a 0.35M solution of Ba3(PO4)2
LekaFEV [45]

Answer:

= 331.81 g

Explanation:

Molarity is calculated by the formula;

Molarity = Moles/volume in liters

Therefore;

Moles = Molarity ×Volume in liters

          = 0.35 M × 1.575 L

          = 0.55125 Moles

But; Molar mass of Ba3(PO4)2 is 601.93 g/mol

Thus;

Mass = 0.55125 moles × 601.93 g/mol

         <u>=331.81 g</u>

7 0
4 years ago
The chemical formula for emerald is Be3Al2(SiO3)6. An emerald can be described as
uysha [10]

Answer:

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Explanation:

6 0
3 years ago
Read 2 more answers
Calculate the enthalpy change, ΔH, for the process in which 30.1 g of water is converted from liquid at 10.1 ∘C to vapor at 25.0
gayaneshka [121]
You can split the process in two parts:

1) heating the liquid water from 10.1 °C to 25.0 °C , and

2) vaporization of liquid water at constant temperature of 25.0 °C.


For the first part, you use the formula ΔH = m*Cs*ΔT

ΔH = 30.1g * 4.18 j/(g°C)*(25.0°C - 10.1°C) = 1,874 J

For the second part, you use the formula ΔH = n*ΔHvap

Where n is the number of moles, which is calculated using the mass and the molar mass of the water:

n = mass / [molar mass] = 30.1 g / 18.0 g/mol = 1.67 mol

=> ΔH = 1.67 mol * 44,000 J / mol = 73,480 J

3) The enthalpy change of the process is the sum of both changes:

ΔH total =  1,874 J + 73,480 J = 75,354 J

Answer: 75,354 J
7 0
3 years ago
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