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3 years ago

If a solution of potassium hydroxide is mixed with sulfuric acid, which type of reaction will most likely occur?

2 answers:

7 0

If a solution of potassium hydroxide is mixed and or chemically reacted with sulphuric acid, the type of reaction that will occur is a Neutralization reaction or a unique type of a Double Replacement Reaction.

Solution = Double Replacement.

k0ka [10]3 years ago

3 0

I think the correct answer from the choices listed above is the last option. If a solution of potassium hydroxide is mixed with sulfuric acid, the type of reaction that will most likely occur would be a double replacement. The reaction is written as:

KOH + H2SO4 = K2SO4 + H2O

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asambeis [7]

Answer:

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2 years ago

Which forces are intramolecular and which intermolecular?

AleksAgata [21]

Butter won't melt in a fridge because of intermolecular tensions. While the bonds inside of the fat molecules are unbroken, the attractions between the fat molecules are weaker.

What intermolecular forces are present in butter?

The intermolecular forces known as London dispersion forces are the weakest and are most prominent in hydrocarbons. Due to the fact that butter molecules are hydrocarbons, London dispersion forces do exist between them.

How do intermolecular forces affect melting?

More energy is required to stop the attraction between these molecules as the intermolecular forces become more powerful. Because of this, rising intermolecular forces are accompanied with rising melting points.

Which forces are intramolecular and which are intermolecular?

Intramolecular forces are those that hold atoms together within molecules. The forces that hold molecules together are known as intermolecular forces.

Learn more about intermolecular forces: brainly.com/question/9328418

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4 0

1 year ago

Consider the titration of 100.0 mL of 0.280 M propanoic acid (Ka = 1.3 ✕ 10−5) with 0.140 M NaOH. Calculate the pH of the result

Murljashka [212]

Answer:

(a) 2.7

(b) 4.44

(d) 5.363

(e) 5.570

(f) 12.30

Explanation:

Here we have the titration of a weak acid with the strong base NaOH. So in part (a) simply calculate the pH of a weak acid ; in the other parts we have to consider that a buffer solution will be present after some of the weak acid reacts completely the strong base producing the conjugate base. We may even arrive to the situation in which all of the acid will be just consumed and have only the weak base present in the solution treating it as the pOH and the pH = 14 -pOH. There is also the possibility that all of the weak base will be consumed and then the NaOH will drive the pH.

Lets call HA propanoic acid and A⁻ its conjugate base,

(a) pH = -log √ (HA) Ka =-log √(0.28 x 1.3 x 10⁻⁵) = 2.7

(b) moles reacted HA = 50 x 10⁻³ L x 0.14 mol/L = 0.007 mol

mol left HA = 0.28 - 0.007 = 0.021

mol A⁻ produced = 0.007

Using the Hasselbalch-Henderson equation for buffer solutions:

pH = pKa + log ((A⁻/)/(HA)) = -log (1.3 x 10⁻⁵) + log (0.007/0.021)= 4.89 + (-0.48) = 4.44

mol HA left = 0.028 -0.014 = 0.014 mol

mol A⁻ produced = 0.014

pH = -log (1.3 x 10⁻⁵) + log (0.014/0.014) = 4.886

(d) mol HA reacted = 150 x 10⁻³ L x x 0.14 mol/L = 0.021 mol

mol HA left = 0.028 - 0.021 = 0.007

mol A⁻ produced = 0.021

pH = -log (1.3 x 10⁻⁵) + log (0.021/0.007) = 5.363

(e) mol HA reacted = 200 x 10⁻³ L x 0.14 mol/L = 0.028 mol

mol HA left = 0

Now we only a weak base present and its pH is given by:

pH = √(kb x (A⁻) where Kb= Kw/Ka

Notice that here we will have to calculate the concentration of A⁻ because we have dilution effects the moment we added to the 100 mL of HA, 200 mL of NaOH 0.14 M. (we did not need to concern ourselves before with this since the volumes cancelled each other in the previous formulas)

mol A⁻ = 0.028 mOl

Vol solution = 100 mL + 200 mL = 300 mL

(A⁻) = 0.028 mol /0.3 L = 0.0093 M

and we also need to calculate the Kb for the weak base:

Kw = 10⁻¹⁴ = ka Kb ⇒ Kb = 10⁻¹⁴/1.3x 10⁻⁵ = 7.7 x 10⁻ ¹⁰

pH = -log (√( 7.7 x 10⁻ ¹⁰ x 0.0093) = 5.570

(f) Treat this part as a calculation of the pH of a strong base

moles of OH = 0.250 L x 0.14 mol = 0.0350 mol

mol OH remaining = 0.035 mol - 0.028 reacted with HA

= 0.007 mol

(OH⁻) = 0.007 mol / 0.350 L = 2.00 x 10 ⁻²

pOH = - log (2.00 x 10⁻²) = 1.70

pH = 14 - 1.70 = 12.30

4 0

2 years ago

What mass of radon is in 8.17 moles of radon?

Len [333]

Answer:

1813.74g

Explanation:

Given parameters:

Number of moles of radon = 8.17moles

Unknown:

Mass of radon = ?

Solution:

To solve this problem, we use the expression below:

Number of moles = $\frac{mass}{molar mass}$

Molar mass of radon = 222g/mol

Now insert the parameters and solve;

Mass of radon = Number of moles x molar mass

= 8.17 x 222

= 1813.74g

4 0

3 years ago

Wich substances is most likely to form in a precipitation reaction?

Flura [38]

Answer:

A precipitation reaction refers to the formation of an insoluble salt when two solutions containing soluble salts are combined. The insoluble salt that falls out of solution is known as the precipitate, hence the reaction's name.

Explanation:

4 0

3 years ago