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What is the density of a liquid with a mass of 31.1415 g and a volume of 30.13 cm³?

Short Answer

Expert verified

Density =1.034 (round four significant figures, Three decimal places)

Step by step solution

01

Calculate the density

It is given that,

Mass of liquid=31.1415g

Volume of liquid=30.13cm3

Density=Mass / Volume

d=31.1415g / 30.13cm3

=1.0335711915 g/cm3

02

Apply the rules of significant figures

All of the digits in measurement, including the uncertain last digit, are called significant figures or significant digits.

Note that zero may be a measured value; for example, if you stand on a scale that shows weight to the nearest pound and it shows "120," then the 1 (hundreds), 2 (tens) and 0 (ones) are all significant (measured) values.

So, rounding off to 4 significant figures, the density will be 1.034 g/cm3

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Most popular questions from this chapter

How much heat is produced when 100 mL of 0.250 M HCl (density, 1.00 g/mL) and 200 mL of 0.150 M NaOH (density, 1.00 g/mL) are mixed?

\({\bf{HCl(aq) + NaOH(aq)}} \to {\bf{NaCl(aq) + }}{{\bf{H}}_{\bf{2}}}{\bf{O(l) \Delta H}}_{{\bf{298}}}^{\bf{^\circ }}{\bf{ = - 58kJ}}\)

If both solutions are at the same temperature and the heat capacity of the products is 4.19 J/g °C, how much will the temperature increase? What assumption did you make in your calculation?

What are the allowed values for each of the four quantum numbers:n,l,ml, andms?

Identify each of the following statements as being most similar to a hypothesis, a law, or a theory. Explain your reasoning.

(a) Falling barometric pressure precedes the onset of bad weather.

(b) All life on earth has evolved from a common, primitive organism through the process of natural selection.

(c) My truck’s gas mileage has dropped significantly, probably because it’s due for a tune-up

Both graphite and diamond burn.

\({\bf{C}}\left( {{\bf{s, diamond}}} \right){\bf{ + }}{{\bf{O}}_{\bf{2}}}\left( {\bf{g}} \right) \to {\bf{C}}{{\bf{O}}_{\bf{2}}}\left( {\bf{g}} \right)\)

For the conversion of graphite to diamond:

\({\bf{C}}\left( {{\bf{s, graphite}}} \right) \to {\bf{C}}\left( {{\bf{s, diamond}}} \right){\bf{ \Delta {\rm H}}}_{{\bf{298}}}^{\bf{^\circ }}{\bf{ = 1}}{\bf{.90 kJ}}\)

Which produces more heat, the combustion of graphite or the combustion of a diamond?

Hydrogen gas, H2, reacts explosively with gaseous chlorine, \({\bf{C}}{{\bf{l}}_{\bf{2}}}\), to form hydrogen chloride, HCl(g). What is the enthalpy change for the reaction of 1 mole of \({{\bf{H}}_{\bf{2}}}\)(g) with 1 mole of \({\bf{C}}{{\bf{l}}_{\bf{2}}}\)(g) if both the reactants and products are at standard state conditions? The standard enthalpy of formation of HCl(g) is −92.3 kJ/mol.

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