They have the same amount of valence electrons
The correction is D. This is because, the higher the value of K, the greater the relative amount of product present at equilibrium.
The molecular formula is D. C_8H_20O_4Si.
<em>Step 1</em>.Calculate the <em>empirical formula
</em>
a) Calculate the moles of each element
Moles of C= 196.01 g C × (1 mol C/12.01 g C) = 16.325 mol C
Moles of H = 41.14 g H × (1 mol H/1.008 g H) = 40.813 mol H
Moles of O = 130.56 g O × (1 mol O/16.00 g O) = 8.1650 mol O
Moles of Si = 57.29 g Si × (1 mol Si/28.085 g Si) = 2.0399 mol Si
b) Calculate the molar ratio of each element
Divide each number by the smallest number of moles and round off to an integer
C:H:O:Si = 8.0027:20.008:4.0027:1 ≈ 8:20:4:1
c) Write the empirical formula
EF = C_8H_20O_4Si
<em>Step </em>2. Calculate the <em>molecular formula</em>
EF Mass = 208.33 u
MF mass = 208.329 u
MF = (EF)_n
n = MF Mass/EF Mass = 208.329 u/208.33 u = 1.0000 ≈ 1
MF = C_8H_20O_4Si
Answer:
= 1.5 eq
Explanation:
One definition of an equivalent weight is that it is mass of a substance that gains or loses 1 mole of electrons.
Al3+ has lost 3 e-, so there are 3 equivalent weights in 1 mol Al3+.
1 mol Al3+ =3 eq. wts.
1 mol Al x(27 g / 1 mol)x(1 mol / 3 eq. wts.) = 9.0 g = 1 eq. wts.
13.5 g Al3 + x (1 eq.wt. / 9.0 g) = 1.5 eq