Answer:
i have same problem bro or sis
Explanation:
C I think it’s C I’m semi guessing
Explanation:
This is correct!
Ions that exist in both the reactant and product side of the equation are referred to as spectator ions. Overall, they do not partake in the reaction. If they are present on both sides of the equation, you can cancel them out.
An example is;
Na+(aq) + Cl−(aq) + Ag+(aq) + NO3−(aq) → Na+(aq) + NO3−(aq) + AgCl(s)
The ions; Na+, NO3−(aq) would be cancelled out to give;
Cl−(aq) + Ag+(aq) → AgCl(s)
Answer:
92.49 %
Explanation:
We first calculate the number of moles n of AgBr in 0.7127 g
n = m/M where M = molar mass of AgBr = 187.77 g/mol and m = mass of AgBr formed = 0.7127 g
n = m/M = 0.7127g/187.77 g/mol = 0.0038 mol
Since 1 mol of Bromide ion Br⁻ forms 1 mol AgBr, number of moles of Br⁻ formed = 0.0038 mol and
From n = m/M
m = nM . Where m = mass of Bromide ion precipitate and M = Molar mass of Bromine = 79.904 g/mol
m = 0.0038 mol × 79.904 g/mol = 0.3036 g
% Br in compound = m₁/m₂ × 100%
m₁ = mass of Br in compound = m = 0.3036 g (Since the same amount of Br in the compound is the same amount in the precipitate.)
m₂ = mass of compound = 0.3283 g
% Br in compound = m₁/m₂ × 100% = 0.3036/0.3283 × 100% = 0.9249 × 100% = 92.49 %
Answer:
The net chemical equation is: 6 H2O(g) + C3H8(g) → 10 H2(g) + 3 CO2(g)
Explanation:
First equation:
The reactants propane and water react to form the products CO and H2
C3H8(g) + 3H2O(g) → 3CO(g) + 7H2(g)
Second equation:
The products CO and H2 will react
CO(g) + H2O(g) → H2(g) + CO2(g)
We should multiply the equation by 3 (to equal the products of the first equation)
3CO(g) + 3H2O(g) → 3H2(g) + 3CO2(g)
Add the second to the first equation:
C3H8(g) + 3 H2O(g) + 3 CO(g) + 3 H2O(g) → 3 H2(g) + 3CO2(g) + 3 CO(g) + 7 H2(g)
The net chemical equation is: 6 H2O(g) + C3H8(g) → 10 H2(g) + 3 CO2(g)
