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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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Solid arsenic reacts with Oxygen gas to synthesize solid arsenic trioxide. How many moles of arsenic trioxide will be produced from 3.4 moles of oxygen gas?
$4As$ + $ 3O_2$ -----> $2As_2O_3$ According to equation $3$ moles of oxygen gas is required to form arsenic trioxide = $2$ moles $3.4$ moles of O2 will be required...
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If a 2.60g sample of titanium metal chemically combines with chlorine gas to form 10.31g of a titanium chloride, what is the empirical formula product?
The idea here is that the difference between the mass of the titanium chloride and the mass of titanium metal will represent the mass of chlorine. This means that...
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What is the empirical formula and empirical formula mass for the following compound: C3H6O3?
By definition, the empirical formula is the SIMPLEST whole number ratio that defines constituent atoms in a species. The ratio of atoms in that species is 1:2:1 with respect to...
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What is the empirical formula for a compound that is 60.9% arsenic and 39.1% sulfur?
How did we get this formula? We assume $100.0$ $g$ of compound, and there are $60.9$ $g$ of arsenic, and $39.1$ $g$ of sulfur. We divide each mass thru...
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What is the empirical formula of a compound that contains 75.7% arsenic and 24.3% oxygen?
Empirical formula is the simplest formula for a compound. Arsenic's symbol is $As$ while oxygen's symbol is $O$ Firstly, we need to make an assumption that the mass of the...
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What is the empirical formula of a substance with 1.587 g arsenic and 3.755 g of chlorine?
In order to find a compound's empirical formula, you must find the smallest whole number ratio that exists between its constituent . In your case, the compound is said...
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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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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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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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