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3 years ago

What is the approximate percent by mass of oxygen in C2H4O2

Chemistry

1 answer:

kherson [118]3 years ago

6 0

Answer:

Approximately 53.3 %.

Explanation:

Molar mass of C2H4O2

= 2 * 12.011 + 4 * 1.008 + 2 * 15.999

= 60.052

% oxygen = (31.998* 100) / 60.052 = 53.28 %

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Aclosed system contains an equimolar mixture of n-pentane and isopentane. Suppose the system is initially all liquid at 120°C an

garri49 [273]

Explanation:

The given data is as follows.

T = $120^{o}C$ = (120 + 273.15)K = 393.15 K,

As it is given that it is an equimolar mixture of n-pentane and isopentane.

So, $x_{1}$ = 0.5 and $x_{2}$ = 0.5

According to the Antoine data, vapor pressure of two components at 393.15 K is as follows.

$p^{sat}_{1}$ (393.15 K) = 9.2 bar

$p^{sat}_{1}$ (393.15 K) = 10.5 bar

Hence, we will calculate the partial pressure of each component as follows.

$p_{1} = x_{1} \times p^{sat}_{1}$

= $0.5 \times 9.2 bar$

= 4.6 bar

and, $p_{2} = x_{2} \times p^{sat}_{2}$

= $0.5 \times 10.5 bar$

= 5.25 bar

Therefore, the bubble pressure will be as follows.

P = $p_{1} + p_{2}$

= 4.6 bar + 5.25 bar

= 9.85 bar

Now, we will calculate the vapor composition as follows.

$y_{1} = \frac{p_{1}}{p}$

= $\frac{4.6}{9.85}$

= 0.467

and, $y_{2} = \frac{p_{2}}{p}$

= $\frac{5.25}{9.85}$

= 0.527

Calculate the dew point as follows.

$y_{1}$ = 0.5, $y_{2}$ = 0.5

$\frac{1}{P} = \sum \frac{y_{1}}{p^{sat}_{1}}$

$\frac{1}{P} = \frac{0.5}{9.2} + \frac{0.5}{10.2}$

$\frac{1}{P}$ = 0.101966 $bar^{-1}$

P = 9.807

Composition of the liquid phase is $x_{i}$ and its formula is as follows.

$x_{i} = \frac{y_{i} \times P}{p^{sat}_{1}}$

= $\frac{0.5 \times 9.807}{9.2}$

= 0.5329

$x_{z} = \frac{y_{i} \times P}{p^{sat}_{1}}$

= $\frac{0.5 \times 9.807}{10.5}$

= 0.467

4 0

3 years ago

How many moles of ammonia, NH,, are contained in<br><br> 6.21 x 1024 molecules of ammonia?

ankoles [38]

Answer:

10.32 moles of ammonia NH₃

Explanation:

From the question given above, the following data were obtained:

Number of molecules = 6.21×10²⁴ molecules

Number of mole of NH₃ =?

The number of mole of NH₃ can be obtained as follow:

From Avogadro's hypothesis,

6.02×10²³ molecules = 1 mole

Therefore,

6.21×10²⁴ molecules = 6.21×10²⁴ / 6.02×10²³

6.21×10²⁴ molecules = 10.32 moles

Thus, 6.21×10²⁴ molecules contains 10.32 moles of ammonia NH₃

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2 years ago

How is an electron orbital similar to a parabola?

AVprozaik [17]

Answer:

Explained below.

Explanation:

First of all, the orbital path of electron is mostly parabolic in electric field.

In an electric field, electrons behave very similar to a projectile. Thus, Electrons have a parabolic path in an electric field simply because the speed of the electrons in a direction which is perpendicular to the electric field is constant since there is no force. Therefore, there will be no acceleration along that perpendicular direction. However there will be an acceleration that is constant in the direction of the electric field which makes it act in a similar manner to a projectile under gravity.

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3 years ago

ADD NOTE

Nikitich [7]

Answer:

2.00 M KOH

Explanation:

Divide moles by volume and make sure your volume is in LITERS, not mL.

0.0400 mol KOH / 0.0200 L = 2.00 M KOH

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3 years ago

How much more acidic is an acid with a pH of 6 than water with a pH of 7?

vladimir1956 [14]

Answer:

1 pH more acidic

Explanation:

0 is considered to be most acidic.

14 is the most alkaline

7 is neutral.

So from 7 pH, it becomes 1 pH acidic, making it 6 pH.

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3 years ago

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