Answer:
Explanation:
Hello,
In this case, considering that the by-mass percent of water is:
Given such percent and the mass of the sample, we can find the mass of water in grams in the sample by solving for it as shown below:
Best regards.
<span>Answer:
It depends on what came after "0.5440 M H...".
If it was a monoprotic acid, like HCl, the calculation would go like this:
(55.25 mL) x (0.5440 M acid) x (1 mol KOH / 1 mol acid) / (0.2450 M KOH) =
122.7 mL KOH
If it was a diprotic acid, like H2SO4, like this:
(55.25 mL) x (0.5440 M acid) x (2 mol KOH / 1 mol acid) / (0.2450 M KOH) =
245.4 mL KOH
If it was a triprotic acid, like H3PO4, like this:
(55.25 mL) x (0.5440 M acid) x (3 mol KOH / 1 mol acid) / (0.2450 M KOH) =
368.0 mL KOH</span>
The final temperature of the water is the equilibrium temperature, or the also the final temperature of the iron after a long period of time. Applying the conservation of energy:
m,iron*C,iron*ΔT = - m,water*C,water*ΔT
The density of water is 1000 g/mL.
(25 g)(0.449 J/g·°C)(T - 398 K) = - (25 mL)(1000 g/mL)(4.18 J/g·°C)(T - 298)
Solving for T,
<em>T = 298.01 K</em>
The graph is needed to answer this question.
Solubility may increase or decrease with temperature depending on the properties of the solute and the solvent.
It is quite common that the solubility of the ionic compounds, like KBr, in water increases with temperature.
Use your solubility curve for the KBr and you wiil see a line that starts at a solubility a little greater than 50 grams of the salt in 100 grams of water for temperaute 0°C and increase linearly until almost 100 grams of the salt in 100 grams of water at 100°C.
So, in this case you can affirm that the solubility of KBr increases with the temperature.
Answer: the second option: the solubility increases.
The question is incomplete , complete question is:
Hydrogen, a potential future fuel, can be produced from carbon (from coal) and steam by the following reaction:
Note that the average bond energy for the breaking of a bond in CO2 is 799 kJ/mol. Use average bond energies to calculate ΔH of reaction for this reaction.
Answer:
The ΔH of the reaction is -626 kJ/mol.
Explanation:
We are given with:
ΔH = (Energies required to break bonds on reactant side) - (Energies released on formation of bonds on product side)
The ΔH of the reaction is -626 kJ/mol.
