The number of grams of Ag2SO4 that could be formed is 31.8 grams
<u><em> calculation</em></u>
Balanced equation is as below
2 AgNO3 (aq) + H2SO4(aq) → Ag2SO4 (s) +2 HNO3 (aq)
- Find the moles of each reactant by use of mole= mass/molar mass formula
that is moles of AgNO3= 34.7 g / 169.87 g/mol= 0.204 moles
moles of H2SO4 = 28.6 g/98 g/mol =0.292 moles
- use the mole ratio to determine the moles of Ag2SO4
that is;
- the mole ratio of AgNo3 : Ag2SO4 is 2:1 therefore the moles of Ag2SO4= 0.204 x1/2=0.102 moles
- The moles ratio of H2SO4 : Ag2SO4 is 1:1 therefore the moles of Ag2SO4 = 0.292 moles
- AgNO3 is the limiting reagent therefore the moles of Ag2SO4 = 0.102 moles
<h3> finally find the mass of Ag2SO4 by use of mass=mole x molar mass formula</h3>
that is 0.102 moles x 311.8 g/mol= 31.8 grams
Answer:
34.6g
Explanation:
Given parameters:
Mass of Octane = 11.2g
Reaction expression;
2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O
Mass of octane = 11.2g
Unknown:
Mass of carbon dioxide produced = ?
Solution:
From the balanced reaction equation;
2 mole of octane produced 16 moles of carbon dioxide
From the given specie, let us find the number of moles;
Number of moles = 
Molar mass of C₈H₁₈ = 8(12) + 18(1) = 114g/mole
Number of moles of octane = 
= 0.098mole
2 mole of octane produced 16 moles of carbon dioxide
0.098 mole of octane will produce 
= 0.79mole of CO₂
Mass of CO₂ = number of moles x molar mass
Molar mass of CO₂ = 12 + 2(16) = 44g/mol
Mass of CO₂ = 0.79 x 44 = 34.6g
Answer:
Heat is a measure of the flow of thermal energy from one object or substance to another. ... Thermal energy typically flows from a warmer material to a cooler material. Generally, when thermal energy is transferred to a material, the motion of its particles speeds up and its temperature increases.
Explanation:
Explanation:
Expresa los gramos de soluto por cada 100 gramos de disolución. Porcentaje masa = masa de soluto___ x 100 masa de la disolución Cuando trabajamos con la masa, podemos sumar el soluto y el disolvente para obtener la disolución.
Answer:
Ecell = +0.25V
Explanation:
the half-cell reactions for a voltanic cell
cathode(reduction): 2H⁺(aq) + 2e⁻ ------- H₂(g)
anode(oxidation): 2AgCl(s) ------- 2Ag⁺(aq) + 2Cl⁻ + 2e⁻
we have the standard cell potential E⁺cell = 0.18V at 80C respectively
Q = [H⁺]/[Cl⁻]
sub for [H+] = 0.10M and [Cl-] = 1.5M
Q= 0.1M/1.5M
Q = 0.067
Ecell = E⁺cell - 
logQ
= 0.18 - 
log 0.067
0.18- 0.059(-1.174)
Ecell = +0.25V
