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Call
Weight of
Weight of
Molecular formula is
Hence, molecular weight =
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Calcium chloride is often used to de-ice roads in the winter. What volume would $15.7$ $g$ calcium chloride occupy, if the density is $2.50$ $g$$/cm^3$?
And so $"volume"="mass"/rho=(15.7*g)/(2.50*g*cm^-3)=6.3*g*cm^3$... And note that is dimensionally consistent given that... $g/(g*cm^-3)=cancelg/(cancelg*1/(cm^3))-=cm^3$...a unit of volume as required....
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The molecular formula of lithium carbonate is Li2CO3. What is the molecular weight of this drug?
As you know this salt is commonly used to treat mania. The formula mass is simply the sum of the molar masses of the contributing atoms: i.e. $(2xx6.94+12.011+3xx15.999)*g*mol^-1$ $=$ $??$
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The recommended daily dose of calcium is 1200 mg. Calcium supplements often use calcium carbonate, $CaCO_3$, as the source of calcium. What mass of $CaCO_3$ should be used in a tablet so that one tablet provides the recommended daily dose of calcium?
$"Moles of calcium"$ $=$ $(1200xx10^-3*g)/(40.08*g*mol^-1)$ $=$ $0.04008*mol$. And thus we need $0.04008*mol$ $"calcium carbonate"$, i.e. a mass of $0.04008*molxx100.09*g*mol^-1$ $=$ $??g$.
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The molar mass of calcium chloride (CaCl2) represents the mass of one mole of what?
The molar mass is the mass of $"Avogadro's number"$ of particles, where $"Avogadro's number "=6.022xx10^23*mol^-1$, and is abbreviated as $N_A$. Thus in one mole of calcium we have $N_A$ calcium...
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Decomposing calcium carbonate yields calcium oxide and dioxide. What information is needed to calculate the mass of calcium oxide that can produced from 4.7 kg of calcium carbonate?
The balanced chemical equation The molar mass of $"CaCO"_3$ The molar ratio of $"CaO"$ to $"CaCO"_3$ The molar mass of $"CaO"$ Your strategy has four steps:...
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How can I balance this chemical equation? Sodium phosphate and calcium chloride react to form calcium phosphate and sodium chloride
Your unbalanced equation is $"Na"_3"PO"_4 + "CaCl"_2 → "Ca"_3("PO"_4)_2 + "NaCl"$ 1. Start with the most complicated formula, $"Ca"_3("PO"_4)_2$. Put a 1 in front of it. We have now...
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Calcium carbonate decomposes to form calcium oxide and carbon dioxide gas. If 44.5 grams of calcium oxide is produced, how many grams of calcium carbonate had to decomposez?
We need a stoichiometric reaction that represents the decomposition of calcium carbonate: $CaCO_3(s) + Delta rarr CaO(s) + CO_2(g)uarr$ And thus calcium oxide and calcium carbonate are present in equimolar...
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A $2.03*g$ mass of calcium salts contains $1.33*g$ calcium carbonate. What is the $%$ calcium carbonate; and what is the percentage by mass of of calcium in calcium carbonate?
So we have the $"%calcium carbonate, "m/m$. We are not finished yet, because we have to address the percentage by mass of calcium metal in calcium carbonate: $%Ca=(40.08*g*mol^-1)/(100.09*g*mol^-1)=40%$ with respect...
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Hydrogen peroxide has the empirical formula HO and an empirical formula weight of 17.0 amu. If the molecular weight is 34.0 amu, what is the molecular formula?
First, you want to find the ratio between the empirical formula and molecular formula; that is, how much has the molecular formula been simplified to reach the empirical formula. We...
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Calcium oxide and carbon dioxide are produced from the decomposition of calcium carbonate. How many moles of calcium carbonate are needed to produce 15.0 grams of the calcium oxide?
We interrogate the stoichiometric equation...... $CaCO_3(s) + Delta rarr CaO(s) + CO_2(g) uarr$ We require a molar quantity of $(15.0*g)/(56.08*g*mol^-1)=0.268*mol,$ with respect to $"calcium oxide"$. And clearly, we need a...
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