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Question 1.1.5 In a series of experiments to determine the density of gold(Au), the results in the second column of the table below were determined. Taking 19.320(g)/(m)L to be the true value for the density of gold, complete the table by using equation (1.1.1) to calculate the absolute error and equation (1.1.2) to calculate the percent error for each density determination. Enter absolute error values in units of (g)/(m)L to the nearest 0.001(g)/(m)L. Include the unit using the symbol (g)/(m)L. Enter percent error values to the nearest 0.01% without units and without the percent symbol. Question1.1.5 In a series of experiments to determine the density of gold (Au),the results in the second column of the table below were determined.Taking 19.320 g/mL to be the true value for the density of gold,complete the table by using eguation1.1.1to calculate the absolute error and eguation1.1.2 to calculate the percent error for each density determination. Enter absolute error values in units of g/mL to the nearest 0.001 g/mL.Include the unit using the symbol g/mL Enter percent error values to the nearest 0.01% without units and without the percent symbol. Results Analysis Absolute Error Determination Au Density Percent Error 1 18.440 g/mL Number Units Number 2 19.427 g/mL Number Units Number 3 19.232 g/mL Number Units Number

          Question 1.1.5
In a series of experiments to determine the density of gold(Au), the results in the second column of the table below were determined. Taking 19.320(g)/(m)L to be the true value for the density of gold, complete the table by using equation (1.1.1) to calculate the absolute error and equation (1.1.2) to calculate the percent error for each density determination.
Enter absolute error values in units of (g)/(m)L to the nearest 0.001(g)/(m)L. Include the unit using the symbol (g)/(m)L. Enter percent error values to the nearest 0.01% without units and without the percent symbol.
Question1.1.5
In a series of experiments to determine the density of gold (Au),the results in the second column of the table below were determined.Taking 19.320 g/mL to be the true value for the density of gold,complete the table by using eguation1.1.1to calculate the absolute error and eguation1.1.2 to calculate the percent error for each density determination. Enter absolute error values in units of g/mL to the nearest 0.001 g/mL.Include the unit using the symbol g/mL Enter percent error values to the nearest 0.01% without units and without the percent symbol.
Results
Analysis Absolute Error
Determination
Au Density
Percent Error
1
18.440 g/mL
Number
Units
Number
2
19.427 g/mL
Number
Units
Number
3
19.232 g/mL
Number
Units
Number
        
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question 115 in a series of experiments to determine the density of goldau the results in the second column of the table below were determined taking 19320gml to be the true value for the de 36936

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Chemistry: Structure and Properties
Chemistry: Structure and Properties
Nivaldo Tro 2nd Edition
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Question 1.1.5 In a series of experiments to determine the density of gold(Au), the results in the second column of the table below were determined. Taking 19.320(g)/(m)L to be the true value for the density of gold, complete the table by using equation (1.1.1) to calculate the absolute error and equation (1.1.2) to calculate the percent error for each density determination. Enter absolute error values in units of (g)/(m)L to the nearest 0.001(g)/(m)L. Include the unit using the symbol (g)/(m)L. Enter percent error values to the nearest 0.01% without units and without the percent symbol. Question1.1.5 In a series of experiments to determine the density of gold (Au),the results in the second column of the table below were determined.Taking 19.320 g/mL to be the true value for the density of gold,complete the table by using eguation1.1.1to calculate the absolute error and eguation1.1.2 to calculate the percent error for each density determination. Enter absolute error values in units of g/mL to the nearest 0.001 g/mL.Include the unit using the symbol g/mL Enter percent error values to the nearest 0.01% without units and without the percent symbol. Results Analysis Absolute Error Determination Au Density Percent Error 1 18.440 g/mL Number Units Number 2 19.427 g/mL Number Units Number 3 19.232 g/mL Number Units Number
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CHM 111 Homework 1 You must write down all of your mathematical steps to get full credit. Please use separate pieces of paper for your answers and please staple them together. Due: Fri 9/20 Question 1 Calculate the density of the following substances. Express your answers in g/cm³ and round up your answers to the nearest significant figures: (a) A 3.00 kg block of chromium displaces 0.4210 L of water. What is the density of chromium? (b) 4.00 x 10⁻² mL of bromine (a liquid element) weigh 0.1249 g. What is the density of bromine? Question 2 Round up your answers to the nearest significant figures. Please use the factor-label method/dimensional analysis format for your calculations. (a) The density of titanium is 4.54 g/cm³. What is the volume of 1.121 g of titanium? (b) The density of germanium is 5.323 g/cm³. How much does a 3.00 x 10³ cm³ block of germanium weigh (in kg)? (c) The density of hexane is 0.6548 g/cm³. How many grams are there is 0.250L of hexane? (d) The density of pentane is 0.626 g/cm³. How many mL are there in 0.450 kg of pentane? (e) The energy content of natural gas is about 39 MJ/m³ (where 1 MJ = 10⁶J). Suppose you need to produce 5.0 x 10⁵J of energy. How much volume (in m³) of natural gas do you need? (f) Sea water has a typical salinity of 39 g/L in dissolved salts. Suppose you have 125.0 mL of sea water. How many grams of dissolved salts can you recover? (g) The atmospheric pressure at the surface of planet Mars is 7.5 mbar (as reported by the Phoenix Mars Mission, http://phoenix.lpl.arizona.edu/). What is the pressure in mmHg (1 bar = 1000 mbar; 750.062 mmHg = 1 bar)? (h) Suppose you have a 75 J/s light bulb (J/s is also call watt, thus it is a 75 watt bulb). Suppose the light bulb was on for 7.200 x 10³ s (i.e. 2 hours). How much energy (in kJ) has it used?

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Transcript

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00:01 Hi, in this question for the first part we need to find out the density.
00:04 We are given mass of object that is 6 .31 grams according to archimedes principle.
00:15 When a solid is immersed in water it displaces water which is equal to its volume.
00:40 So, here volume will be equal to volume of water displaced that is 23 .7 minus 19 .1 it comes out 4 .6 milliliters.
00:53 Now volume of the solid plastic phase is 4 .6 milliliter density can be calculated as mass divided by volume on dividing mass 6 .31 grams with 4 .6 milliliter we get density as 1 .37 gram per milliliter.
01:16 Now moving to the next part we have mass of 25 pannies which is 64 .72 grams and water displaced by 25 pannies is equal to 31 .54 minus 23 .41 milliliter we get 8 .13 milliliter...
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