Which has not been suggested as a reasonably practical way to store large amounts of hydrogen in relatively small spaces for its
1 answer:
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Answer:
The specific heat of the alloy
Explanation:
Mass of an alloy = 25 gm
Initial temperature = 100°c = 373 K
Mass of water = 90 gm
Initial temperature of water = 25.32 °c = 298.32 K
Final temperature = 27.18 °c = 300.18 K
From energy balance equation
Heat lost by alloy = Heat gain by water
[
-
] =
(
-
)
25 × × ( 373 - 300.18 ) = 90 × 4.2 (300.18 - 298.32)
This is the specific heat of the alloy.
Answer:
Upper F subscript 2 (g) plus upper C a (s) right arrow with delta above upper C a upper F subscript 2 (s).
Explanation:
This is a chemical reaction problem.
In expressing any chemical reaction, we need to understand that there are reactants and products.
- The reactants are the species on the left hand side that are combining.
- The products are the species on the right hand side that are formed.
- Every chemical reaction is obeys the law of conservation of matter i.e equal number of matter on both sides.
Using the statement of this problem, we can deduce that;
Reactants are Fluorine gas and Calcium metal
Product is Calcium Fluoride
Note: A metal is a solid(s) and powder is a solid(s). A gas is denoted as (g). They depict the state of the species reacting.
F₂ + Ca
→ CaF₂
We can see that equal number of atoms are on both sides of the expression.
Answer : The correct option is, (A) -101.37 KJ
Explanation :
First we have to calculate the moles of HCl and NaOH.
The balanced chemical reaction will be,
From the balanced reaction we conclude that,
As, 1 mole of HCl neutralizes by 1 mole of NaOH
So, 0.1392 mole of HCl neutralizes by 0.1392 mole of NaOH
Thus, the number of neutralized moles = 0.1392 mole
Now we have to calculate the mass of water.
As we know that the density of water is 1 g/ml. So, the mass of water will be:
The volume of water =
Now we have to calculate the heat absorbed during the reaction.
where,
q = heat absorbed = ?
= specific heat of water =
m = mass of water = 174 g
= final temperature of water = 317.4 K
= initial temperature of metal = 298 K
Now put all the given values in the above formula, we get:
Thus, the heat released during the neutralization = -14.11 KJ
Now we have to calculate the enthalpy of neutralization.
where,
= enthalpy of neutralization = ?
q = heat released = -14.11 KJ
n = number of moles used in neutralization = 0.1392 mole
Therefore, the enthalpy of neutralization is, -101.37 KJ