Answer:
2C3H18 + 15O2 ---->6CO2 +18H2O
Explanation:
the number of reactant must be equal to the number of product.
Answer:
15.95
Explanation:
This question is a modification of the calculation of the empirical formula of a compound given its percent composition and atomic weights of the elements in the compound.
Here we are given the formula and the percent composition, so we know that there are 4 atoms of E per 2 atoms of N so lets solve using the information given.
In 100 grams of the binary compound we have
30.46 g N
69.54 g E
The number of moles is the mass divided by atomic weight:
mol N = 30.46 g / A.W N = 30.46 g / 14.00 g/mol = 2.18 mol N
mol E = 65.54 g / A.W E
Thus,
4 mol E/ 2 mol N = ( 69.54 g/ A.W E ) / 2.18
2 A.E = 65.54 g / 2.18 ⇒ A.W E = 69.54 g / ( 2 x 2.18 ) = 15.94 g
So the A.W is 15.94 g/mol which is close the atomic weight of O.
Answer:
Ksp = 8.8x10⁻⁵
Explanation:
<em>Full question is:</em>
<em>After mixing an excess PbCl2 with a fixed amount of water, it is found that the equilibrium concentration of Pb2+ is 2.8 × 10–2 M. What is Ksp for PbCl2?</em>
<em />
When an excess of PbCl₂ is added to water, Pb²⁺ and Cl⁻ ions are produced following Ksp equilibrium:
PbCl₂(s) ⇄ Pb²⁺ + 2Cl⁻
Ksp = [Pb²⁺] [Cl⁻]²
If an excess of PbCl₂ was added, an amount of Pb²⁺ is produced (X) and twice Pb²⁺ is produced as Cl⁻ (2X):
Ksp = [X] [2X]²
Ksp = 4X³
As X is the amount of Pb²⁺ = 2.8x10⁻²M:
Ksp = 4(2.8x10⁻²)³
<h3>Ksp = 8.8x10⁻⁵</h3>
Answer:
Es la tercera por qué cuando el material varía va cambiando de estado
Explanation:
