Binary are compounds that were made from two different .
They have a formula of
Some examples include:
$Na + Cl -> NaCl$ => Sodium chloride
$2P + 3H_"2" -> 2PH_"3"$ => Phosphorus trihydride
$2Sn + O_"2" => 2SnO$ => Tin (II) oxide
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Alternatively, $"copper (I) selenide"$. Copper here is in the univalent oxidation state.
1 Answers 1 viewsThis is also known as $"quick lime"$ or $"burnt lime"$.
1 Answers 1 viewsThe compound, nitrogen(IV) oxide, is a radical species as written (it has 17 valence electrons), and participates in the following equilibrium: $2NO_2(g) rightleftharpoonsN_2O_4(g)$ The dimer is colourless, and contains a...
1 Answers 1 viewsStep 1. Calculate the empirical formula. The empirical formula is the simplest whole-number ratio of atoms in a compound. The ratio of atoms is the same as the ratio of...
1 Answers 1 viewsAs always with these problems, it is useful to assume $100*g$ of unknown compound....and thus find an empirical formula... $"Moles of carbon"=(12.8*g)/(12.011*g*mol^-1)=1.066*mol.$ $"Moles of hydrogen"=(2.1*g)/(1.00794*g*mol^-1)=2.084*mol.$ $"Moles of bromine"=(85.1*g)/(79.90*g*mol^-1)=1.066*mol.$ And we...
1 Answers 1 viewsAs is typical with these questions, we assume $100*g$ of unknown compound, and work out the MOLAR quantities of each element present: $" moles of C":$ $(15.8*g)/(12.011*g*mol^-1)=1.32*mol$ $" moles of...
1 Answers 1 viewsWe can write an equation for the (complete) combustion of the substance as follows: $C_xH_yN_z + (x+y/2+z)/2O_2 to xCO_2+y/2H_2O+zNO_2$ That might look complicated, but all we're saying is that each...
1 Answers 1 viewsIn $100* g$ of this there are $30.4* g$ $N$, and the balance $O$. We divide thru by the atomic masses in order to approach the empirical formula: $(30.4*g)/(14.01*g*mol^-1)$ $=$...
1 Answers 1 viewsThe of tin in the compound will be equal to the ratio between the mass of tin and the mass of the compound, multiplied by $100$. $color(blue)("% Sn"...
1 Answers 1 viewsFirst, work out the mass of each element that was present in the original compound. Carbon is always present as $CO_2$ in the ratio (12.011 g / 44.0098 g), and...
1 Answers 1 views