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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 reacts with chlorine to form the compound sodium $2Na + Cl_2 -> 2NaCI$. How would you describe, in terms of electron arrangement, the type of bonding in a molecule of chlorine?
The chlorine molecule is represented as $Cl-Cl$, i.e. $Cl_2$. Between the chlorine atoms, 2 electrons overlap to form a region of high electron to which the positively charged chlorine nuclei...
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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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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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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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A chemist performs the synthesis of sodium chloride from its elements. The chemist begins with 46 grams of sodium. How many moles of chlorine are needed?
2 Na + Cl2 → 2 NaCl
$"Moles of sodium "=(46.0*g)/(22.9*g*mol^-1)=2*mol$. Given your equation, clearly we need the ONE mole of $Cl_2$ gas. What is the mass of this quantity of chlorine? It is fact that all...
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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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Sodium and chlorine react to form sodium chloride: 2Na (s) + Cl2 (g) ----> 2NaCl (s)
For the reaction of 44.0g Na with 69.9 g Cl2, what is the limiting reactant and what is the theoretical yield of sodium chloride in grams?
This is a limiting reactant problem. We know that we will need a balanced equation with masses, molar masses, and moles of the involved. 1. Gather all the information...
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When ethane, $C_2H_6$, reacts with chlorine gas the main product is $C_2H_5Cl$ but small amounts of $C_2H_4Cl_2$ are produced. What is the percent yield of $C_2H_5Cl$ if the reaction of 125g of ethane with 255g of chlorine gas produced 206g of $C2H_5Cl$?
The key here is to focus solely on the reaction that produces chloroethane, $"C"_2"H"_5"Cl"$, and ignore the one that produces dichloroethane, $"C"_2"H"_4"Cl"$, the side product of the reaction. Start by...
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