1. start with balanced equation.
2 H2(g) + O2(g<span>) </span><span> 2 H</span>2O(g<span>)
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2. Use stoichiometry
(3.4moles of H)(2moles of H2O/2moles of H)
The moles of H will cancel, leaving you with moles of H2O.
The answer is 3.4 moles of H2O
Answer:
6.24 x 10-3 M
Explanation:
Hello,
In this case, for the given dissociation, we have the following equilibrium expression in terms of the law of mass action:
![Ka=\frac{[H_3O^+][BrO^-]}{[HBrO]}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BH_3O%5E%2B%5D%5BBrO%5E-%5D%7D%7B%5BHBrO%5D%7D)
Of course, water is excluded as it is liquid and the concentration of aqueous species should be considered only. In such a way, in terms of the change 
, we rewrite the expression considering an ICE table and the initial concentration of HBrO that is 0.749 M:
Thus, we obtain a quadratic equation whose solution is:
Clearly, the solution is 0.00624 M as no negative concentrations are allowed, so the concentration of BrO⁻ is 6.24 x 10-3 M.
Best regards.
Answer:
T2 = 29°C
Explanation:
Given data:
Heat added = 420 j
Mass of water = 25 g
Initial temperature = 25°C
Final temperature = ?
Solution;
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Specific heat capacity of water = 4.18 j/g.°C
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
Now we will put the values.
420 j = 25 g ×4.18 j/g.°C × (Final temperature - initial temperature)
420 j = 25 g ×4.18 j/g.°C × (T2 - 25°C)
420 j = 104.5 j/°C × (T2 - 25°C)
420 j /104.5 j/°C = T2 - 25°C
4°C + 25°C = T2
T2 = 29°C
Answer:
2Au₂S₃ + 6H₂ → 4Au + 6H₂S
Explanation:
Balancing:
2Au₂S₃ + 6H₂ → 4Au + 6H₂S
