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Specific Gravity Calculator: Need to solve the specific gravity of a material in fraction of seconds? Direct link to Mark Zwald's post All objects have a buoyan, Posted 5 years ago. would be 2.7, which means that If multiplying, I come out with 472.298 liters. You can target the Engineering ToolBox by using AdWords Managed Placements. Specific Gravity. )%2F01%253A_Matter_and_Measurements%2F1.12%253A_Density_and_Specific_Gravity, \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\), status page at https://status.libretexts.org, Ice \(\left( 0^\text{o} \text{C} \right)\), Water \(\left( 4^\text{o} \text{C} \right)\). 154 ml 1. Since most materials expand as temperature increases, the density of a substance is temperature dependent and usually decreases as temperature increases. A is denser than B, so A would sink into B or B would float on A. would be submerged beneath the water's surface. WebThe conversion formula used by this tool is: SG = / 0 Symbols = Density of substance in kgm -3 0 = Density of pure water (1000 kgm -3 @4C) SG = Specific gravity (e.g. The completed calculation is: [16.2 g sol'n x [37.3 g HCl /100 g sol'n] / [36.47 g HCl/ -).3 z d/HFs/i:X9 .Q :I!`j{B ett }B; 0@%@4klT(dqq +fR@20 M Q-b`66(Sh )l7i)`G1010fdrcXhw20401U0a:$EaGe1,ap*4C,2olx&(L%@ r For this to work, the density must be defined using the same units of mass and volume as referenced in the rest of the formula. gravity of an object is the density of Specific Gravity The differences in intermolecular interactions, the density of substance can change at different rates compared to the density of water, that's why the specific gravity also changes with changing pressure and temperature.