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Dmitry [639]
2 years ago
7

The organic compound di-n-butyl phthalate, C16H22O4(l), is sometimes used as a low-density (1.046 g·mL–1) manometer fluid. Compu

te the pressure (in torr) of a gas that supports a 535-mm column of di-n-butyl phthalate. The density of mercury is 13.53 g·mL–1.
Chemistry
2 answers:
solniwko [45]2 years ago
8 0

Answer:

P=(mRT/MV) then P=dRT/M

so at STP:

T= 25 celsius which equals 298.73 Kelvin

d= 1.046 g/mL = 0.001046 g/L

R =0.08206 Latm/Kmol

M= molar mass of C16H22O4, which is (12)16+(1)22+(16)4 = 278 g/mol

P=dRT/M

1 atm = 760 torr = 760 mm Hg

multiply P by 760

Explanation:

olganol [36]2 years ago
4 0
The correct answer to this question is this one:

use PV=nRT

we can convert it into P=(mRT/MV) then P=dRT/M
so at STP:
T= 25 celsius which equals 298.73 Kelvin
<span />d= 1.046 g/mL = 0.001046 g/L (we need to use litres in the ideal gas equation)
<span />R =0.08206 Latm/Kmol
<span />M= molar mass of C16H22O4, which is (12)16+(1)22+(16)4 = 278 g/mol


<span>so plug this into: P=dRT/M

once you found the pressure, P, you need to convert it into torr (because it will be in atm (atmospheres).. since 1 atm = 760 torr = 760 mm Hg

multiply P by 760 and you're done. You;ll have the pressure in torr </span>
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Answer:

0.3229 M HBr(aq)

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Explanation:

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<em />

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

<em>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?</em>

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

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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

6 0
2 years ago
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Answer:

59.2 grams

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Explanation:

Credit goes to @znk

Hope it helps you :))

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