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2 years ago

What is the mass percentage of O in C₂H₄O₂? Provide an answer to two decimal places.

Chemistry

1 answer:

nasty-shy [4]2 years ago

6 0

Answer:

53.29%

Explanation:

The molar mass of C2H4O2 is 60.05g and the 2 O's are 32.00g

so 32.00/60.05= 0.53288925895

and that as a decimal rounded to the nearest hundredths is 53.29%

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Formation of water: 2H2 + 1 O2 --> 2H2O

noname [10]

Answer:

2.2 moles H2O

Explanation:

$35g O_2 \mbox{ \cdot }\frac{1mol}{32g/mol} \mbox{ \cdot }\frac{2mol H_2O}{1mol O_2}= 2.1875$ , which rounds to about 2.2

7 0

2 years ago

What is formed when hydrobromic acid, hbr, and calcium hydroxide, ca(oh)2, are combined?

Lorico [155]

This is an acid – base reaction and this always result a salt and water in a neutralization reaction. 

The salt that is formed will be calcium bromide (calcium is located in group 2 so calcium bromide has a formula of CaBr2)

so essentially we got:

HBr + Ca(OH)2 ------> CaBr2 + H2O

balancing the elements: 

2HBr(aq) + Ca(OH)2(aq) --------> CaBr2(aq) + 2H2O(l)

6 0

3 years ago

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

The most abundant gases in the atmosphere are _____. trace gases variable gases greenhouse gases primary gases

White raven [17]

The correct answer is greenhouse gases. It is the most abundant gases among the choices in the atmosphere. These gases are water vapor, methane, nitrous oxide, ozone and carbon dioxide. Without these gases, the temperature of Earth will be about -18 degrees Celsius.

3 0

3 years ago

Read 2 more answers

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vichka [17]

Light moves in a straight line except at surfaces between different transparent materials, where its path bends.

4 0

3 years ago