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Mumz [18]
3 years ago
6

How does the molecular formula of a compound differ from the empirical formula? Can a compound’s empirical and molecular formula

s be the same? Explain.
Chemistry
1 answer:
Y_Kistochka [10]3 years ago
4 0
Molecular is every element present in the compound eg C2H6, empirical is the smallest whole number ratio of elements in a compound so that would be CH3 as you divide by the highest common factor. Some compounds only have 1 formula if they are simple or have no common factors. Eg methane, CH4 is its molecular and empirical because its the simplest whole number ratio and includes every element in the molecule
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3 0
3 years ago
Why are organic compounds predominantly covalent?
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5 0
3 years ago
For example, if the electronegativity of both atom "a" and "b" in the phet are set to be equivalent, then increasing the electro
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8 0
3 years ago
Read 2 more answers
Determine the rate law and the value of k for the following reaction using the data provided: 2N2O5(g) → 4 NO2(g) + O2(g)0.2241.
BartSMP [9]
2N2O5(g)----> 4NO2(g) + O2(g) 
<span>[N2O5]i (M) Initial Rate(M^-1 s^-1) </span>
<span>0.093 4.84x10^-4 ---- (1) </span>
<span>0.186 9.67x10^-4 ----- (2) </span>
<span>0.279 1.45x10^-3 ----- (3) </span>
<span>From equation (1) & (2) it is evident that when [N2O5}i is doubled the initial rate is doubled, which implies the rate is directly proportional to [N2O5]. Similarly comparing equation (1) & (3) we observe that when [N2O5] is tripled the rate is also tripled. Hence the rate equation is </span>
<span>Rate = k [N2O5] </span>
<span>Using the data of any equation, say (1), we get </span>
<span>4.84x10^-4 = k x 0.093 </span>
<span>OR k = 4.84x10^-4/0.093 = 5.2 x 10^-3 s-1 </span>
<span>Hence the rate law is </span>
<span>Rate = 5.2 x 10^-3 s-1[N2O5]</span>
5 0
3 years ago
Read 2 more answers
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