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erica [24]
3 years ago
6

Volume measurements will be made using a 25-ml graduated cylinder, which has major scale divisions marked every 2 ml and minor s

cale divisions every 0.2 ml. what is the estimated uncertainty in the volume measurements
Chemistry
2 answers:
KATRIN_1 [288]3 years ago
7 0

Answer:

The estimated uncertainty is ±0.1 ml

Explanation:

Measurements are usually associated with uncertainty which depends on the instrument used or the procedure employed.

It is given that the graduated cylinder of volume 25 ml has a major scale marked every 2 ml and minor scale with 0.2 ml. In general measurements made with graduated cylinders can be estimated one decimal place beyond the smallest division. In this case, the volume of the liquid can be read to the nearest 0.1 division. Hence, the uncertainty would be ± 0.1 ml.

Free_Kalibri [48]3 years ago
5 0

Actually, the uncertainty point is when we have to estimate one place smaller than what the graduated cylinder can accurately read. So in this case, the smallest that we can accurately read is given by the minor marks of 0.2 mL. So if it is in between two minor marks we can only estimate the number, therefore the uncertainty is +/- 0.1ml.

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Which is greater 50 degree celcius or 50 degree fahrenheit?with solution plz.....​
zheka24 [161]

Answer: 50 degrees Celsius is hotter than 50 degrees F. 50 degrees Celsius is half way between freezing and boiling

4 0
3 years ago
Which of the following shows how rate depends on concentrations of reactants?
Katarina [22]

Answer:

Rate = k(A)m(B)n

Explanation:

A.P.E.X.

6 0
3 years ago
8. A sample of sulfur has a mass of 223 g. How many moles of sulfur are in the sample?
Rama09 [41]

Answer:

7 moles

Explanation:

1mole of sulfur=32

x=223

=6.9

=7moles

8 0
3 years ago
Combustion vapor-air mixtures are flammable over a limited range of concentrations. The minimum volume % of vapor that gives a c
DochEvi [55]
Combustion equation of n-hexane:

2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O

a)
Assuming we have 100 moles of air,
Oxygen = 20.9 moles
n-hexane required = 20.9/19 x 2
= 2.2 moles

LFL = Half of stoichometric amount = 2.2 / 2 = 1.1
LFL n-hexane = 1.1% 

b)
1.1 volume percent required for LFL

1.1% x 1
= 0.0011 m³ of n-hexane required
6 0
3 years ago
Read 2 more answers
A bomb calorimeter has a heat capacity of 675 J/°C and contains 925 g of water. If the combustion of 0.500 mole of a hydrocarbon
ikadub [295]

<u>Answer:</u> The enthalpy of the reaction is 269.4 kJ/mol

<u>Explanation:</u>

To calculate the heat absorbed by the calorimeter, we use the equation:

q_1=c\Delta T

where,

q = heat absorbed

c = heat capacity of calorimeter = 675 J/°C

\Delta T = change in temperature = T_2-T_1=(53.88-24.26)^oC=29.62^oC

Putting values in above equation, we get:

q_1=675J/^oC\times 29.62^oC=19993.5J

To calculate the heat absorbed by water, we use the equation:

q_2=mc\Delta T

where,

q = heat absorbed

m = mass of water = 925 g

c = heat capacity of water = 4.186 J/g°C

\Delta T = change in temperature = T_2-T_1=(53.88-24.26)^oC=29.62^oC

Putting values in above equation, we get:

q_2=925g\times 4.186J/g^oC\times 29.62^oC=114690.12J

Total heat absorbed = q_1+q_2

Total heat absorbed = [19993.5+114690.12]J=134683.62J=134.7kJ

To calculate the enthalpy change of the reaction, we use the equation:

\Delta H_{rxn}=\frac{q}{n}

where,

q = amount of heat absorbed = 134.7 kJ

n = number of moles of hydrocarbon = 0.500 moles

\Delta H_{rxn} = enthalpy change of the reaction

Putting values in above equation, we get:

\Delta H_{rxn}=\frac{134.7kJ}{0.500mol}=269.4kJ/mol

Hence, the enthalpy of the reaction is 269.4 kJ/mol

6 0
3 years ago
Read 2 more answers
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