The density of mercury metal is 13.6 g/mL. What is the mass 45 mL of the metal?

There are $0.454*lb*(kg^-1)$ so $2.203*kg*(lb^-1)$, and $1*cm^3 = 1*mL$ So we just do the conversion: $5.0*cancel(L)xx1000*cancel(cm^3)cancel(L^-1)xx13.6*g*cancel(cm^-3)$ $=$ $??g$; ; $??cancel(g)xx10^(-3)cancel(kg)*cancel(g^-1)xx(2.203)lb*cancel(kg^-1)$ $=$ $??$ $lb$

We need to (i) find the volume of mercury; and (ii) convert this volume into a mass by use of the quoted . $"Volume of mercury"$ $=$ $76.0*cmxxpixx(0.5*cm)^2$ $=$ $59.7$...

For a given substance, its tells you the mass occupied by one unit of volume of that substance. In essence, is a measure of how well the molecules of...

A substance's basically tells you how much mass of said substance occupies one unit of volume. In your case, the of mercury is given in Grams per milliliter,...

The density, $rho$, of mercury $=$ $13,593*kg*m^-3$; it is $13.6*g*cm^-3$ in more accessible units. Anyway, mercury metal is pretty dense stuff. If you ever found a pool of mercury, you...

By definition, $"density,"$ $rho="mass"/"volume"$... ..and thus to get the $"mass"$ we take the product.... $"mass"=rhoxx"volume"$...and here are quoted both $rho$ and $"volume"$. And so..... $"mass"=underbrace(13.6*g*cancel(cm^-3))_"density"xxunderbrace(16.8*cancel(cm^3))_"volume"=underbrace(??*g)_"mass"$. Often, in the lab you...

By definition, $rho,"density"$ $=$ $"Mass of stuff"/"Volume of stuff"$. We want to find a volume, so: $"Volume"$ $=$ $"Mass"/rho$ $=$ $(175*cancelg)/(13.6*cancelg*mL^-1)$. Note that this calculation is dimensionally consistent. We want...

Using the formula $Density=(Mass)/(Volume)$, we can rearrange it as $Mass=Density*Volume$. The volume given is 29.6mL and the given is 13.6g/ml. Plugging it into our equation, we get $Mass=29.6cancel(ml)*(13.6g)/cancel(ml)=402.56g$

Ionic bonds occur between anions and cations. And cations may be metallic, i.e., $M^(2+)$, or they may be non-metallic, i.e. ammonium salts, $NH_4^+$. And anions, while typically non-metallic, i.e. $X^-$...

The problem tells you that the sample is $7.00%$ mercury by mass, which automatically implies that it is $93.00%$ aluminium by mass. That is the case because the amalgam...