The balanced chemical equation for the above reaction is as follows;
2LiOH + H₂SO₄ ---> Li₂SO₄ + 2H₂O
stoichiometry of base to acid is 2:1
Number of OH⁻ moles reacted = number of H⁺ moles reacted at neutralisation
Number of LiOH moles reacted = 0.400 M / 1000 mL/L x 20.0 mL = 0.008 mol
number of H₂SO₄ moles reacted - 0.008 mol /2 = 0.004 mol
Number of H₂SO₄ moles in 1 L - 0.500 M
This means that 0.500 mol in 1 L solution
Therefore 0.004 mol in - 1/0.500 x 0.004 = 0.008 L
therefore volume of acid required = 8 mL
Answer:
Ammonium bromide can be prepared by the direct action of hydrogen bromide on ammonia. It can also be prepared by the reaction of ammonia with iron(II) bromide or iron(III) bromide, which may be obtained by passing aqueous bromine solution over iron filings.
Explanation:
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The reaction of iron (III) oxide and aluminum is initiated by heat released from a small amount "starter mixture". This reaction is an oxidation-reduction reaction, a single replacement reaction, producing great quantities of heat (flame and sparks) and a stream of molten iron and aluminum oxide which pours out of a hole in the bottom of the pot into sand.
The balanced chemical equation for this reaction is:
2 Al(s) + Fe2O3(s) --> 2Fe(s) + Al2O3(s) + 850 kJ/mol
Curriculum Notes
This chemical reaction can be used to demonstrate an exothermic reaction, a single replacement or oxidation-reduction reaction, and the connection between ∆H calculated for this reaction using heats of formation and Hess' Law and calculating ∆H for this reaction using qrxn = mc∆T and the moles of limiting reactant. This reaction also illustrates the role of activation energy in a chemical reaction. The thermite mixture must be raised to a high temperature before it will react.
To determine how much thermal energy is released in this reaction, heats of formation values and Hess' Law can be used.
By definition, the deltaHfo of an element in its standard state is zero.
2 Al(s) + Fe2O3(s) --> 2Fe (s) + Al2O3 (s)
The deltaH for this reaction is the sum of the deltaHfo's of the products - the sum of the deltaHfo's of the reactants (multiplying each by their stoichiometric coefficient in the balanced reaction equation), i.e.:
deltaHorxn = (1 mol)(deltaHfoAl2O3) + (2 mol)(deltaHfoFe) - (1 mol)(deltaHfoFe2O3) - (2 mol)(deltaHfoAl)
deltaHorxn = (1 mol)(-1,669.8 kJ/mol) + (2 mol)(0) - (1 mol)(-822.2 kJ/mol) - (2mol)(0 kJ/mol)
deltaHorxn = -847.6 kJ
The melting point of iron is 1530°C (or 2790°F).
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Answer: 1) Temperature can change the solubility of a solute.
Explanation:
The chart is missing so there is no way to tell what does the graph show.
Yet, I can help you because I can explain the status of each statement of the choices. As you will see there is only one possibility..
<span>1) Temperature can change the solubility of a solute.
Yes, temperature definetly can, and mostly do, modify the solubility of a solute.
You can search any chart of solubility and will find that.
I can give you two examples:
a) Sodium chloride: dissolve some spoons of salt in a cold water until you can not dissolve more. Then, heat the water, you will find that more salt will get dissolved, proving that the temperature of the solution increases the solubility of sodium chloride.
b) Carbon dioxide gas: the soft drinks have CO₂ molecules dissolved in it.
The higher the temperature of the soft drink the less the amount of CO₂(g) that can be dissolved. That is why the soda bottling plants cool the beverage before adding the CO₂(g).
2) </span><span>Temperature has no affect on the solubility of a solute.
Since this is the opposite to the first statement and the first is true, this is false.
3) Salt has a greater solubility than sugar.
False.
This is an empirical result, which you cannot predict theoretically. So you need to see at the data either in a table or in a chart. Else you can test it at home. After the empirical data are shown it results that more grams of sugar can be dissolved in water compared to salt.
That is something you ca see in a chart or you can prove by yourself.
4) Nitrite salt has a greater solubility than sugar.
</span>
False.
Looking at some data you can find that sodium nitrite solutiliby is aroun 70 - 100 g/10 g while sugar (sucrose) solutiblity is around 180 - 235 g/ 100 g.
