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2 years ago

The indicator propyl red has ka = 3.3 × 10–6 . estimate the approximate ph range over which this indicator would change color.

1 answer:

7 0

We let the chemical equation for the weak acid indicator propyl red HPr be
HPr + H2O ↔ H3O+ + Pr-
Therefore, the acid dissociation constant Ka is
Ka = [H3O+][Pr-] / [HPr]
The color of this indicator turns from red to yellow or the other way around at its turning point at which
[HPr] = [Pr-]
Substituting this to the equation for Ka, we now have
Ka = [H3O+]
The pH of the solution at its turning point is
pH = -log[H3O+] = -log(Ka) = -log 3.3×10^-6 = 5.48
On this account, the pH range is pH 5.0 to 6.0.

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What will be the charge of the ion formed from each of these atoms ?

Galina-37 [17]

Answer:

Si14- Si^4+

As33- As^3-

Mg12- Mg^2+

Rb37- Rb^+

F9- F^-

Ge32- Ge^4+

Sn50- Sn^2+, Sn^4+

Explanation:

The elements shown in the answer have their common ions written beside them.

Silicon mostly forms positive ions in oxyacids and complex ions. Arsenic mostly forms its anion. Magnesium forms only the +2cation just as rubidium only forms the +1 cation. The fluoride ion is F^- while tin may for a +2*or +4 cation. Germanium usually forms the +4 cation.

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2 years ago

An increase in temperature will effect vapor pressure by:

-BARSIC- [3]

Answer: Increases.

Explanation: As the temperature of a liquid or solid increases its vapor pressure also increases. Conversely, vapor pressure decreases as the temperature decreases.

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2 years ago

When C2H5Cl(g) is burned in oxygen, chlorine gas is produced in addition to carbon dioxide and water vapor. 5145 kJ of heat are

Arisa [49]

Answer: The chemical equation is written below.

Explanation:

Combustion reaction is defined as the chemical reaction in which a hydrocarbon reacts with oxygen gas to produce carbon dioxide gas and water molecule.

$\text{hydrocarbon}+O_2\rightarrow CO_2+H_2O$

The chemical equation for the combustion of ethyl chloride follows:

$4C_2H_5Cl+13O_2\rightarrow 2Cl_2+8CO_2+10H_2O$

We are given:

When 4 moles of ethyl chloride is burnt, 5145 kJ of heat is released.

For an endothermic reaction, heat is getting absorbed during a chemical reaction and is written on the reactant side.

$A+\text{heat}\rightleftharpoons B$

For an exothermic reaction, heat is getting released during a chemical reaction and is written on the product side

$A\rightleftharpoons B+\text{heat}$

So, the chemical equation follows:

$4C_2H_5Cl+13O_2\rightarrow 2Cl_2+8CO_2+10H_2O+5145kJ$

Hence, the chemical equation is written above.

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2 years ago

Stu Dent has finished his titration, and he comes to you for help with the calculations. He tells you that 20.00 mL of unknown c

Alexus [3.1K]

Answer:

0.3229 M HBr(aq)

0.08436M H₂SO₄(aq)

Explanation:

Stu Dent has finished his titration, and he comes to you for help with the calculations. He tells you that 20.00 mL of unknown concentration HBr(aq) required 18.45 mL of 0.3500 M NaOH(aq) to neutralize it, to the point where thymol blue indicator changed from pale yellow to very pale blue. Calculate the concentration (molarity) of Stu's HBr(aq) sample.

Let's consider the balanced equation for the reaction between HBr(aq) and NaOH(aq).

NaOH(aq) + HBr(aq) ⇄ NaBr(aq) + H₂O(l)

When the neutralization is complete, all the HBr present reacts with NaOH in a 1:1 molar ratio.

$18.45 \times 10^{-3} L NaOH.\frac{0.3500molNaOH}{1LNaOH} .\frac{1molHBr}{1molNaOH} .\frac{1}{20.00 \times 10^{-3} LHBr} =\frac{0.3229molHBr}{1LHBr} =0.3229M$

Kemmi Major also does a titration. She measures 25.00 mL of unknown concentration H₂SO₄(aq) and titrates it with 0.1000 M NaOH(aq). When she has added 42.18 mL of the base, her phenolphthalein indicator turns light pink. What is the concentration (molarity) of Kemmi's H₂SO₄(aq) sample?

Let's consider the balanced equation for the reaction between H₂SO₄(aq) and NaOH(aq).

2 NaOH(aq) + H₂SO₄(aq) ⇄ Na₂SO₄(aq) + 2 H₂O(l)

When the neutralization is complete, all the H₂SO₄ present reacts with NaOH in a 1:2 molar ratio.

$42.18 \times 10^{-3} LNaOH.\frac{0.1000molNaOH}{1LNaOH} .\frac{1molH_{2}SO_{4}}{2molNaOH} .\frac{1}{25.00\times 10^{-3}LH_{2}SO_{4}} =\frac{0.08436molH_{2}SO_{4}}{1LH_{2}SO_{4}} =0.08436M$

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2 years ago

How many molecules are in 79g of fe2o3

horrorfan [7]

Convert grams —> mols and then mols —> atoms

We know that there are 6.02 x 10^23 atoms/mol

And we know that there are about 160 grams of fe2o3 per mol

So (79g fe2o3)/(160 g/mol) = .49 mol fe2o3

Now we use avogadro’s number to do

(.49 mol fe2o3)/(6.02 x 10^23 atoms/mol) = the answer.

I’ll leave the easy math to you.

7 0

2 years ago