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slega [8]
3 years ago
8

When phenolphthalein indicator is added to a colorless solution with a pH of 10, a student observes and concludes that the teste

d solution
1
remains colorless and is basic
2.
remains colorless and is acidic
3.
turns pink and is basic
4.
turns pink and is acidic
Chemistry
1 answer:
madam [21]3 years ago
4 0

Answer:

3.  turns pink and is basic

Explanation:

Phenolphthalein is a pH indicator that remains colorless in acidic solutions, but in basic solutions it turns pink at a pH equal to 10.

Phenolphthalein is a weak acid that loses H+ cations in solution. The phenolphthalein molecule is colorless, while the phenolphthalein-derived anion is pink. When a base is added, phenolphthalein loses H+, forming the anion and causing it to turn pink. The color change cannot be explained only on the basis of deprotonation, a structural change occurs with the appearance of a ketoenolic tautomerism.

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What is the identity of a gas that has a density of 1.4975 g/L and a volume of 8.64 L at a pressure of 2.384 atm and with a temp
bagirrra123 [75]
Answer: H2O (water)

Explanation:

The answer choices for this question are:

A) H2O
B) N2
C) SO2
D) NO3
E) Cl2

The solution of the problem is:

1) Data:

<span> density, d = 1.4975 g/liter
volume, V = 8.64 liter
pressure, p = 2.384 atm
temperature, T = 349.6 K

2) Formulas:

d = m/V => m = d*V

n = m / molar mass => molar mass = m / n

pV = nRT => n = pV / RT

3) Solution

n = pV / RT = 2.384 atm * 8.64 liter / (0.0821 atm*liter/K*mol * 349.6K)

n = 0.7176 moles

</span>m = dV = 1.4975 g/ liter * <span>8.64 liter = 12.9384 g

molar mass = m / n = 12.9384 g / 0.7176 moles = 18.03 g/mol

That molar mass corresponds to the molar mass of water, therefore the gas is H2O (water vapor).</span>
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8 0
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Read 2 more answers
A power plant is driven by the combustion of a complex fossil fuel having the formula C11H7S. Assume the air supply is composed
AlekseyPX

(a) 4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 20.68N_2;

(b) 4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2;

(c) 23 900 kg air; (d) air:fuel = 10.2; (e) air:fuel = 12.2:1

(a) <em>Balanced equation including N_2 from air</em>  

The balanced equation <em>ignoring</em> N_2 from air is  

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2  

Moles of N_2 =55 mol O_2 × (3.76 mol N_2/1 mol O_2) = 206.8 mol N_2  

<em>Including</em> N_2 from air, the balanced equation is  

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 206.8N_2  

(b) <em>Balanced equation for 120 % stoichiometric combustion</em>  

Moles of O_2 = 55 mol O_2 × 1.20 = 66.00 mol O_2  

Excess moles O_2 = (66.00 – 55) mol O_2 = 11.00 mol O_2  

Moles of N_2 = 66.00 mol O_2 × (3.76 mol N_2/1 mol O_2) = 248.2 mol N_2  

The balanced equation is

4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2

(c) <em>Minimum mass of air</em>  

Moles of O_2 required = 1700 kg C_11 H_7S

× (1 kmol C_11 H_7S/185.24 kg C_11 H_7S) × (55 kmol O_2/4 kmol C_11 H_7S)

= 126.2 kmol O_2  

Mass of O_2 = 126.2 kmol O_2 × (32.00 kg O_2/1 kmol O_2) = 4038 kg O_2  

Mass of N_2 required = 126.2 kmol O_2 × (3.76 kmol N_2/1 kmol O_2)

× (28.01 kg N_2/1 kmol N_2) = 13 285 kg N_2  

Mass of air = Mass of N_2 + mass of O_2 = (4038 + 13 285) kg = 17 300 kg air  

(d) <em>Air:fuel mass ratio for 100 % combustion</em>  

Air:fuel = 17 300 kg/1700 kg = <em>10.2 :1 </em>

(e) <em>Air:fuel mass ratio for 120 % combustion </em>

Mass of air = 17 300 kg × 1.20 = 20 760 kg air  

Air:fuel = 20 760 kg/1700 kg = 12.2 :1  

6 0
3 years ago
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