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Elanso [62]
4 years ago
10

Nitrogen gas is introduced into a large deflated plastic bag. No gas is allowed to escape, but as more and more nitrogen is adde

d, the bag inflates to accommodate it. The pressure of the gas within the bag remains at 1.00 atm and its temperature remains at room temperature (20.0∘C). How many moles n have been introduced into the bag by the time its volume reaches 22.4 L
Chemistry
1 answer:
Natalija [7]4 years ago
4 0

Answer:

0.9307 moles have been introduced into the bag.

Explanation:

Pressure of the gas within the bag,P = 1.00 atm

Temperature of the gas remains at room temperature,T=20.0 °C = 293.15 K

Volume of the gas in the bag = V = 22.4 L

Number of moles of gas = n

Using an ideal gas equation:

PV=nRT

1.00 atm\times 22.4 atm=n\times 0.0821 atm l/mol K\times 293.15 K

n = 0.9307 moles

0.9307 moles have been introduced into the bag.

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An analytical chemist is titrating of a solution of nitrous acid with a solution of . The of nitrous acid is . Calculate the pH
Burka [1]

Answer:

pH = 2.69

Explanation:

The complete question is:<em> An analytical chemist is titrating 182.2 mL of a 1.200 M solution of nitrous acid (HNO2) with a solution of 0.8400 M KOH. The pKa of nitrous acid is 3.35. Calculate the pH of the acid solution after the chemist has added 46.44 mL of the KOH solution to it.</em>

<em />

The reaction of HNO₂ with KOH is:

HNO₂ + KOH → NO₂⁻ + H₂O + K⁺

Moles of HNO₂ and KOH that react are:

HNO₂ = 0.1822L × (1.200mol / L) = <em>0.21864 moles HNO₂</em>

KOH = 0.04644L × (0.8400mol / L) = <em>0.0390 moles KOH</em>

That means after the reaction, moles of HNO₂ and NO₂⁻ after the reaction are:

NO₂⁻ = 0.03900 moles KOH = moles NO₂⁻

HNO₂ = 0.21864 moles HNO₂ - 0.03900 moles = 0.17964 moles HNO₂

It is possible to find the pH of this buffer (<em>Mixture of a weak acid, HNO₂ with the conjugate base, NO₂⁻), </em>using H-H equation for this system:

pH = pKa + log₁₀ [NO₂⁻] / [HNO₂]

pH = 3.35 + log₁₀ [0.03900mol] / [0.17964mol]

<h3>pH = 2.69</h3>
8 0
4 years ago
What is thr pH of a solution of NaCl that has [H+] = 1.00 × 10-7 M
almond37 [142]
Ph= - log [H+] = -log 1.00× 10-7 = -(log 1 + log 10-7) = -( 0 + (-7log 10) = -( -7) = 7
3 0
3 years ago
Read 2 more answers
How many grams of NaCl are in 3.5 mol of NaCl? Also, how many moles of NaCl are in 150 g of NaCl?​
alexandr402 [8]

Answer:

204.5505 grams

2.5666 moles

Explanation:

For the first question, multiply 3.5 (# of moles) by 58.443 (g/mol for NaCl).

58.443 * 3.5

<em>I'll distribute 3.5 into 58.443.</em>

(3.5 * 50) + (3.5 * 8) + (3.5 * 0.4) + (3.5 * 0.04) + (3.5 * 0.003)

175 + 28 + 1.4 + 0.14 + 0.0105

203 + 1.4 + 0.14 + 0.0105

204.4 + 0.14 + 0.0105

204.54 + 0.0105

204.5505 grams

There are 204.5505 grams in 3.5 moles of NaCl.

For the second question, divide 150 (# of grams) by 58.443 (g/mol for NaCl). I'll convert both into fractions.

150/1 * 1000/58443

150000/58443

2.56660336 moles

2.5666 moles (rounded to 4 places to keep consistency with the first answer) are in 150 grams of NaCl.

6 0
3 years ago
Type the correct answer in each box. Spell all words correctly.
Dmitry_Shevchenko [17]
There’s nothing there?
8 0
3 years ago
The triple point of a substance is the temperature and pressure conditions at which Select one: a. density is greatest. b. state
N76 [4]

Answer:

Option b.

The triple point of a substance is the temperature and pressure conditions at which states of a substance coexist at equilibrium

3 0
4 years ago
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