Consider the following reversible reaction. 2H2O(g)<—>2H2(g)+O2(g) What is the equilibrium constant expression for the giv
2 answers:
Answer : The equilibrium constant expression will be,
Explanation :
The given balanced chemical reaction is,
The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.
The concentrations of pure solids are constant that is they do not change. Thus, they are not included in the equilibrium expression.
So, the equilibrium constant expression will be,
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The fuel released 90 calories of heat.
Let suppose that water experiments an entirely <em>sensible</em> heating. Hence, the heat released by the fuel is equal to the heat <em>absorbed</em> by the water because of principle of energy conservation. The heat <em>released</em> by the fuel is expressed by the following formula:
(1)
Where:
- Mass of the sample, in grams.
- Specific heat of water, in calories per gram-degree Celsius.
- Temperature change, in degrees Celsius.
If we know that ,
and
, then the heat released by the fuel is:
The fuel released 90 calories of heat.
We kindly invite to check this question on sensible heat: brainly.com/question/11325154
Answer:
The volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml
Explanation:
Here we have the reaction of AgNO₃ and NaCl as follows;
AgNO₃(aq) + NaCl(aq)→ AgCl(s) + NaNO₃(aq)
Therefore, one mole of silver nitrate, AgNO₃, reacts with one mole of sodium chloride, NaCl, to produce one mole of silver choride, AgCl, and one mole of sodium nitrate, NaNO₃,
Therefore, since 5.84 grams of NaCl which is 58.44 g/mol, contains
0.09993 moles of NaCl will react with 0.09993 moles of AgNO₃
Also, as 1.0 M solution of AgNO₃ contains 1 mole per 1 liter or 1000 mL, therefore, the volume of AgNO₃ that will contain 0.09993 moles is given as follows;
0.09993 × 1 Liter/mole= 0.09993 L = 99.93 mL
Therefore, the volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml.