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An average sample of chlorine contains 75.77% chlorine-35, with an atomic mass of 34.969 amu (atomic mass units), and 24.23% chlorine-37, with an atomic mass of 36.966 amu. What is the average atomic mass of chlorine?
The weighted average is the sum of each individual nuclide, multiplied by its isotopic abundance. So, we simply do the arithmetic: $0.7577xx34.969+0.2423xx36.966 = ???$ amu Then look at to...
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Sodium ions bond with chlorine ions to form table salt, NaCl. Why does one sodium ion bond with one chlorine ion?
And of course, in an ionic solid (say $"sodium chloride"$), an isolated cation, a sodium ion, is BOUND ELECTROSTATICALLY to EVERY OTHER CHLORIDE ANION in the lattice. The attraction/repulsion follows...
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When an atom of chlorine forms an ionic bond with an atom of sodium, what happens to the atom of chlorine?
The sodium atom loses its one valence electron to form a cation with a charge of +1, $Na^+$. Chlorine gains the electron to obtain a full valence shell and forms...
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Potassium is a very flammable metal if it comes in to contact with water. As it burns with the water it creates potassium hydroxide ($KOH$) If you separate the potassium out from 50 grams of $KOH$, how many grams of potassium would you have?
$"Moles of potassium hydroxide"$ $=$ $(50*g)/(56.11*g*mol^-1)$ $"Mass of potassium metal"$ $=$ $(50*g)/(56.11*g*mol^-1)xx39.10*g*mol^-1$ $~=$ $??g$
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Which option contains a $2*g$ mass of sodium atoms....
$1.$ $5.24xx10^22*"sodium atoms?"$
$2.$ $6.022xx10^24*"sodium atoms?"$
$3.$ $3.12xx10^23*"sodium atoms?"$
$4.$ $5.24xx10^22*"sodium atoms?"$
We know that $1*mol$ contains $6.022xx10^23$ individual sodium atoms, and has a mass of $22.99*g$. And thus for $1.$ there is a $2*g$ mass of $Na$. For $2.$ there is...
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There are two compounds of titanium and chlorine. One compound contains 31.04% titanium by mass, and the other contains 74.76% chlorine by mass. What are the ratios of titanium and chlorine atoms in the two compounds?
The idea here is that you need to use the of the two in the two to find their . Once you have their , you also have the ratio...
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A $22.35*g$ mass of an unknown compound consisting of potassium, chlorine, and oxygen, contained $7.13*g$ of potassium, $6.47*g$ of chlorine. What is its empirical formula?
As always, we divide each constitiuent mass thru by the of each constituent atom. $"Moles of potassium"$ $=$ $(7.13*g)/(39.10*g*mol^-1)$ $=$ $0.182*mol$. $"Moles of chlorine"$ $=$ $(6.47*g)/(35.45*g*mol^-1)$ $=$ $0.182*mol$. $"Moles...
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Chlorine reacts with hot sodium hydroxide solution. A sample of a compound formed in the reaction was found to contain 10.8 g of sodium atoms, 16.7 g of chlorine atoms and 22.5 g of oxygen atoms. What is the empirical formula of the compound formed?
$"Moles of chlorine"=(16.7*g)/(35.45*g*mol^-1)=0.47*mol$ $"Moles of sodium"=(10.8*g)/(22.99*g*mol^-1)=0.47*mol$ $"Moles of oxygen"=(22.5*g)/(16.00*g*mol^-1)=1.41*mol$ We divide thru by the smallest quantity, that of sodium, to give an empirical formula.....of $NaClO_3$, i.e. $"sodium chlorate"$.
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Recall that liquid sodium reacts with chlorine gas to produce sodium chloride. You want to produce 581 g of sodium chloride. How many grams of sodium are needed?
$"Moles of sodium chloride"$ $(581*cancelg)/(58.44*cancelg*mol^-1)$ $=$ $"approx. "10*mol$ Thus $10$ $mol$ sodium metal are required.
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Sodium and chlorine gas react to produce sodium chloride. If $"234 g"$ of sodium chloride is produced, how much sodium reacted?
You know that $2"Na"_ ((s)) + "Cl"_ (2(g)) -> 2"NaCl"_ ((s))$ and that the reaction produced $"234 g"$ of sodium chloride. Convert this to moles by using the...
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