Answer:
<em>(H30+)= 1x10^-6 M</em>
Explanation:
Both pH and pOH have a relationship to belonging to the same aqueous solution: the expression of the Kwater (ionic product of the water Kw) is used:
1x 10-8 mol/L equals to1x10-8 M
(H3O+) x (OH-) = 1x10^-14
(H30+)x 1x 10^-8 =1x10^-14
(H30+)= 1x10^-14/1x 10^-8
<em>(H30+)= 1x10^-6 M</em>
Respuesta:
90.0 %
Explicación:
Paso 1: Escribir la ecuación química balanceada
N₂ + 3 H₂ ⇒ 2 NH₃
Paso 2: Calcular el rendimiento teórico de NH₃ a partir de 140 g de N₂
En la ecuación balanceada, participan de N₂: 1 mol × 28.01 g/mol = 28.01 g y de NH₃: 2 mol × 17.03 g/mol = 34.06 g.
140 g N₂ × 34.06 g NH₃ /28.01 g N₂ = 170 g NH₃
Paso 3: Calcular el rendimiento porcentual de NH₃
El rendimiento experimental de NH₃ es 153 g. Podemos calcular el rendimiento porcentual usando la siguiente fórmula.
R% = rendimiento experimental / rendimiento teórico × 100%
R% = 153 g / 170 g × 100% = 90.0 %
In determining the boiling point
of solutions, always take note of the number of ions that will dissociate in
the solution. It does not depend on the nature of the substance. The greater
the number of ions dissociated, the greater is the boiling point of the solution.
Answer:
The mass of sugar used is 427.5 grams.
Explanation:
The formula of common sugar is 
.
Its molar mass of molecule is 342 grams per mole.
The number of moles of sugar used is equal to '1.25'
(NOTE : <em>You can calculate molar mass by adding the mass of all the atoms present in the atom. </em>
<em>Mass of Oxygen (O) = 16 ; Total number of Oxygen atoms present = 11</em>
<em>Mass of Hydrogen (H) =1 ; Total number of Hydrogen Atoms present = 22</em>
<em>Mass of Carbon (C) = 12 ; Total number of Carbon atoms present = 12</em>
<em>Total mass of molecule = 12 × 12 + 1 × 22 + 16 × 11 = 342</em>)
