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Nesterboy [21]
3 years ago
11

1.Compare and contrast the rate of solution formation between the three physical forms of salt that were placed in the vial and

not agitated with the three forms of salt that were placed in the vial and were agitated.
2. On the basis of your results, what is the relationship between the temperature of the solvent and the rate of solution formation?
3. Use your knowledge of collision theory to explain the results of your experiments in this laboratory.
Chemistry
1 answer:
lawyer [7]3 years ago
8 0

<span>The stable form of salt at standard temperature and pressure is solid. If you add water to salt, it will dissolve. However, there are certain factors that affect the rate of solution formation of salt. The rate of formation of salt solution is faster when the vial is agitated than when it is not agitated. By agitating the solution, you are increasing the surface area of the salt particles in contact with water.</span>

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Determine the standard enthalpy of formation in kJ/mol for NO given the following information about the formation of NO2 under s
Zarrin [17]

Answer:

90.3 kJ/mol

Explanation:

Let's consider the following thermochemical equation.

2 NO(g) + O₂(g) → 2 NO₂(g)  ∆H°rxn = –114.2 kJ

We can find the standard enthalpy of formation for NO using the following expression.

∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g)) - 1 mol × ΔH°f(O₂(g))

∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g)) - 1 mol × 0 kJ/mol

∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g))

ΔH°f(NO(g)) = (2 mol × ΔH°f(NO₂(g)) - ∆H°rxn) / 2 mol

ΔH°f(NO(g)) = (2 mol × 33.2 kJ/mol + 114.2 kJ) / 2 mol

ΔH°f(NO(g)) = 90.3 kJ/mol

8 0
3 years ago
Balanced chemical equations show:
lara31 [8.8K]

the products formed from the reaction

all of the above

4 0
3 years ago
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Answer:

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6 0
3 years ago
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In a sulphuric acid (h2so4) - sodium hydroxide (naoh) acid-base titration, 17.3 ml of 0.126 m naoh is needed to neutralize 25 ml
katen-ka-za [31]
The balanced equation for the neutralisation reaction is as follows
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry of NaOH to H₂SO₄ is 2:1
the number of moles of NaOH reacted  - 0.126 mol/L  x 0.0173 L = 0.00218 mol
if 2 mol of NaOH reacts with 1 mol of H₂SO₄ 
then 0.00218 mol of NaOH reacts with - 0.00218 / 2 = 0.00109 mol of H₂SO₄ 
molarity is the number of moles of solute in 1 L solution
therefore if 25 mL contains - 0.00109 mol 
then 1000 mL contains - 0.00109 mol / 25 mL  x 1000 mL = 0.0436 mol/L
therefore molarity of H₂SO₄ is 0.0436 M
4 0
3 years ago
Sulfur burns in the air to form 33.6L of sulfur dioxide. (S = 32, O =16, 64g of sulfur dioxide has 22.4L, 32g of oxygen has 22.4
ivolga24 [154]

The mass of oxygen reacted/required in this reaction is obtained as 48g.

<h3>What is stoichiometry?</h3>

The term stoichiometry has to do with mass- volume or mass - mole relationship which ultimately depends on the balanced reaction equation.

Now, we have the reaction; S + O2 ------>SO2

If 1 mole of sulfur dioxide contains 22.4 L

x moles of sulfur dioxide contains 33.6L

x = 1.5 moles of sulfur dioxide.

Since the reaction is 1:1, the number if moles of oxygen required/reacted is 1.5 moles.

Mass of oxygen required/reacted = 1.5 moles * 32 g/mol = 48g

Learn more anout stoichiometry: brainly.com/question/9743981

8 0
2 years ago
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