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A block of wood is floating in water. The weight of the part of the block above water is one third of the total weight of the block. What is the the part the water by the block of weight of wood, using Archimedes' principle?
The question is not clear but I think this is likely what you are looking for. If $2/3$ of the block of wood is submerged then $color(white)("XXX")$the mass of water...
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$rho_"glycerol"=1.20*g*mL^-1$. What volume is associated with a $43.7*g$ mass?
$"Density, "rho$ $=$ $"Mass"/"Volume"$ Thus $"Volume"$ $=$ $"Mass"/rho$ $=$ $(43.7*g)/(1.20*g*mL^-1)$ $=$ $??mL$. Note that this answer is consistent dimensionally. We were asked to find a volume. We got an answer...
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Given $rho_"gold"=19.3*g*cm^-3$, will a thief set off a mass-activated alarm if he steals a gold cylinder which is $22*cm$ long, and which has a diameter of $7.6*cm$, and replaces the gold bar with a bag of sand whose mass is $3*kg$?
The question is asking you to use , $rho$, which is $"mass"/"volume"$, and usually has units of $g*cm^-3$ or $g*mL^-1$ in order to determine the mass of each substance. Since...
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A beaker whose mass was $345.8*g$ was filled with a $225*mL$ volume of carbon tetrachloride, to give a combined mass of $703.55*g$. Given these data, what is $rho_"carbon tetrachloride"$?
And so we put the given numbers in: $rho_"carbon tetrachloride"=(703.55*g-345.8*g)/(225*mL)=??g*mL^-1$ Because $1*g*mL^-1=1xx10^-3kg*(10^-3L)^-1=1xx10^-3kgxx10^3*L^-1=kg*L^-1$, the $"density"$ has the same figures in $kg*L^-1$. $"Carbon tetrachloride"$ will form an immiscible layer with water. Will...
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A $42*kg$ mass of substance occupies a volume of $22*m^3$...what is $rho_"material"$ in $g*L^-1$?
$"Density,"$ $rho$ $=$ $"Mass"/"Volume"$. Chemists generally quote $rho$ in terms of $g*cm^-3$, or $g*L^-1$. And thus $rho=(42xx10^3*g)/(22*m^3xx10^3*L*m^-3)=1.91*g*L^-1$ I think the gas (and how did I know it was a gas?)...
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Given $rho_"substance"=2.78*g*mL^-1$, what volume is associated with a $32.0*g$ mass?
I write the defining quotient because given two of the three properties, we can access the third by division or multiplication..... Here we gots $rho$, and we gots $"mass"$.... And...
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How many moles of aluminium cations are present in $"5.1 g"$ of aluminium oxide?
The idea here is that each formula unit of aluminium oxide, $"Al"_color(red)(2)"O"_color(blue)(3)$, consists of two aluminium cations, $color(red)(2) xx "Al"^(3+)$ three oxygen anions, $color(blue)(3) xx "O"^(2-)$...
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The molar mass of fluorine is 19.00 g/mol and that of aluminium is 26.98 g/mol. How do you calculate the molar mass of aluminium fluoride?
The formula of aluminium fluoride is $"AlF"_3$. This tells us that one formula unit of $"AlF"_3$ contains one atom of $"Al"$ and three atoms of $"F"$. It also...
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How many aluminium atoms are present in 5.396 grams of aluminium?
2Al + 6HCI -> 2AlCl3 + 3H2
We need (i) to calculate the molar quantity: $(5.396*g)/(26.982*g*mol^-1)=0.2000*mol$ And then (ii), multiply this molar quantity by $"Avogadro's number"$, $6.022xx10^23*mol^-1$ to get................................... $0.2*molxx6.022xx10^23=??$
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A hydrate containing aluminium sulphate has the formula $"Al"_2 ("SO"_4)_3 * x"H"_2 "O"$ and it contains 11.11% of aluminium by mass. Calculate the value of $x$ in the hydrate formula.
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Your goal here is to figure out exactly how many moles of water of hydration are present for every mole of hydrate. As you can see, the chemical formula of...
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