$19.3 g/cancel(cm^3)*1.47cancel(cm^3)~~28.4g$
1 Answers 1 viewsWe can use the to solve this problem. $color(blue)(|bar(ul(color(white)(a/a)PV = nRT color(white)(a/a)|)))" "$ Since $"moles" = "mass"/"molar mass"$ or $n = m/M$, we can write $PV...
1 Answers 1 viewsAssuming that the neckless is made from $"24K"$, $24$ karat gold, i. e. gold that is $99.9%$ pure, you can solve this problem by using the mass of the neckless...
1 Answers 1 viewsGold has a , $rho=19.3*g*mL^-1,$ which is approx. 20 times as much water. The density of the gold crown (given to the tyrant of Syracuse, not the king), was...
1 Answers 1 views$"Density, "rho$ $-=$ $"Unit mass"/"Unit Volume"$ Here, both mass and volume were quoted, and we simply take the quotient, knowing that $1*L$ $=$ $1000*cm^3$. Thus $"rho$ $-=$ $(3.15xx10^3g)/(2.50*cancelLxx10^3*cm^3*cancel(L^-1))$ $=$ $??g*cm^-3$
1 Answers 1 views$"Density"$ $=$ $"Mass"/"Volume"$ I have an idea that the of silver metal is greater than $10.0*g*mL^-1$ (this is typical of the heavier metals!). You will have to look this up...
1 Answers 1 viewsd = $m/v$ First fill in the m and v $(18.9g)/(1.12mL)$ Then, divide $(18.9g)/(1.12mL) = 16.88 g//mL$ Thus, making the ring not pure.
1 Answers 1 viewsA pure substance has definite physical and chemical properties....which include melting point, boiling point, (under given conditions),......... And examples of pure substances include all the , water, methane gas, sugar,...
1 Answers 1 viewsIf there are two an element $"Average atomic mass" = "mass"_axx "fract"_a + "mass"_b xx "fract"_b$ In the given problem about Hafnium, 5 isotopes have been given. Extending the...
1 Answers 1 viewsThe $"Average Atomic Mass"$ of an element is defined as "the weighted average mass of all naturally-occurring (occasionally radioactive) of the element." (and hence the name "average") [1] Dividing the...
1 Answers 1 views