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Cloud [144]
3 years ago
13

Choose the appropriate volumetric device to use for each of the volume measuring applications.

Chemistry
1 answer:
Oksanka [162]3 years ago
4 0

Answer:

1. Beaker and Flask: These two volumetric devices are used to make coarse measurements of volume. These devices have graduated volume measurements marked on the side if the beaker and flask. The beaker and flask have a 5% accuracy.

2. Graduated Cylinder: More accurate than beakers and flask, with a 1% accuracy. Graduated cylinder are transparent with finely divided graduations. Their size vary from 5mL to 200mL.

3. Burets: These are cylindrical glassware with a valve at the bottom (commonly called a “stopcock”). They also have graduations on the side for accurate reading, and they have a 0.01mL accuracy.

4. Pipets: 12-24 inches long slender tubes, that measure predetermined volume. Mohr pipets are volumetric glassware pipets with graduations used to measure odd or fractional volumes to be provided. They have a 0.02mL accuracy. There are three types of pipets: air-displacements; positive-displacement; and the dispenser pipet.  

Explanation:

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Identify the true statement(s). There may be one answer or more than one answer.FADH2 is a reducing agent. FADH2 is an oxidizing
adell [148]

Answer:

FADH2 is a reducing agent.

FAD is an oxidizing agent.

Explanation:

The full form of FAD is flavin adenine dinucleotide. It is mainly a redox-active coenzyme which is associated with the different proteins and is involved with the enzymatic reactions in the metabolism.

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Thus we see that FAD is an oxidizing agent while $FADH_2$ is a reducing agent.      

8 0
3 years ago
What is the volume of 2.1 moles of chlorine gas (Cl2) at standard temperature and pressure (STP)?
uysha [10]
Given:
<span> 2.1 moles of chlorine gas (Cl2) at standard temperature and pressure (STP)
Required:
volume of CL2
Solution:
Use the ideal gas law
PV = nRT
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5 0
3 years ago
A student calculated the density of a sample of graphite to be 2.3 g/cm3. Show a numerical setup for calculating the student’s p
LUCKY_DIMON [66]

Answer:

Percent error = 1.5%

Explanation:

Given data:

Measured value of density of graphite = 2.3 g/cm³

Percent error = ?

Solution:

Formula:

Percent error = [Measured value - Actual value / actual value] × 100

Actual/accepted value of density of graphite = 2.266 g/cm³

Now we will put the values:

Percent error = [2.3 g/cm³ - 2.266 g/cm³ / 2.266 g/cm³] × 100

Percent error = [0.034 g/cm³ / 2.266 g/cm³] × 100

Percent error = 0.015  × 100

Percent error = 1.5%

8 0
3 years ago
Read 2 more answers
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