Analyze the table and determine which student reported the most precise set of d

Question

Analyze the table and determine which student reported the most precise set of data. Analyze the table and determine which student reported the most accurate set of data. The actual mass is 5.50 g. A student reports three trials to determine the density of an article and records the data as shown in table below. The accepted value of density is 2.55 g/mL. Calculate the percent error in each trial. Calculate the area of a rectangular piece of metal with a length of 8.0 times 10^5 cm and a width of 4.0 times 10^3 cm. Express your answer in meters and in scientific notation with the proper amount of significant figures. Convert the following temperatures: 350.0 K to degree C then to degree F -80.0 degree F to degree C then to K Solve the following problems using scientific notation rules for multiplication and division. (5 times 10^-11)(1.2 times 10^6)/(5 times 10^4)(6 times 10^-7) = (9.2 times 10^-6)(6.0 times 10^4)/(1.6 times 10^5)(2.0 times 10^-5) =

Explanation / Answer

Answer – 21) We know accuracy refer the how close a measurement is to its standard value. The precision refers to how close two or more measurements are to each other and they are whether accurate or not.

We are given the actual mass is 5.50 g

a) Student A data mass are very close to each other and not accurate, so the student A data is precision

b) Student data B is close to the standard value that is 5.50 value, so student data B is accurate.

22) We are given the accepted value = 2.55 g/mL

We also given the student data with three trails

We know, the formula for percent error is                      

Percent error = | accepted value - measured value | / accepted value * 100 %

For trial 1-

Percent error = | 2.55-2.54| / 2.55 * 100 %

                      = 0.392 %

For trial 2-

Percent error = | 2.55-2.56| / 2.55 * 100 %

                      = 0.392 %

For trial 3-

Percent error = | 2.55-2.57| / 2.55 * 100 %

                      = 0.784 %

23) We are given length = 8.0*105 cm , width = 4.0*103 cm

We need to convert the length and width cm to m

Length -

1 cm = 0.01 m

So, 8.0*105 cm = ?

= 8.0*103 m            

Width -

1 cm = 0.01 m

So, 4.0*103 cm = ?

= 40 m

We know, area of rectangular = length * width

                                                = 8.0*103 m * 40 m

                                                 = 3.2*105 m2

area of rectangular piece is 3.2*105 m2

24) a) We are given temperature in Kelvin and need to convert the temperature is degree Celsius and degree Fahrenheit.

We know,

Temperature in degree Celsius = temperature in Kelvin – 273.15

                                                  = 350.0 K – 273.15

                                                  = 76.85 oC

Temperature in degree Fahrenheit = 9/5 * temp in degree Celsius + 32 )

                                                      = 9/5 * 76.85 oC + 32

                                                      = 170.33 oF

b) Tof = - 80.0oF

we know,

ToC = ( Tof -32) *5/9

       = ( -80.0 -32) *5/9

     = -62.2 oC

We know,

T K = ToC + 273.15

        = -62.2 +273.15

        = 210.95 K

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