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3 years ago

What does a coefficient represent in a chemical formula?

Chemistry

1 answer:

lisov135 [29]3 years ago

8 0

Answer:

Theoretical number of moles

Explanation:

The coefficient used in balancing a chemical reaction helps to show theoretically the number of moles of reactants used or products formed. It tends to give a representation of the ratio of the number of moles of reactants and products.

What we are saying in essence is that it helps to picture the number of moles of the reactants coming into the chemical reaction or the number of moles of products formed as the reaction proceeds to completion

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Plz help me woth these 2 questions they r killing me!!!!!

Bogdan [553]

If you leave the plant in the closet it will die due to the lack of sunlight and water it needs.

Yes, you could begin to swear because you are used to hearing those words so in your mind it will get implanted and it will be just a natural response.

Happy studying ^_^

3 0

4 years ago

a helium balloon has a volume of 2.95 liters at 25 c. the volume of the ballon decreased to 2.25 l after it is placed outside on

OleMash [197]

Answer:

The outside temperature is -45.8°C

Explanation:

When a gas keeps on constant its moles and its pressure, we can assume that volume will be increased or decreased as the T° (absolute T° in K).

V1 / T1 = V2 / T2

2.95L/298K = 2.25L / T2

(2.95L/298K ) . T2 = 2.25L

T2 = 2.25L . 298K / 2.95L

T2 = 227.2K

T°K - 273 = T°C

227.2K - 273 = -45.8°C

3 0

4 years ago

1) How many Joules of energy are released when 75g of water is heated

Pie

Answer: 23,199 J

Explanation:

8 0

2 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

4 years ago

Wet steam at 1100 kPa expands at constant enthalpy (as in a throttling process) to 101.33 kPa, where its temperature is 105°C. W

nikklg [1K]

Answer:

There are 3 steps of this problem.

Explanation:

Step 1.

Wet steam at 1100 kPa expands at constant enthalpy to 101.33 kPa, where its temperature is 105°C.

Step 2.

Enthalpy of saturated liquid Haq = 781.124 J/g

Enthalpy of saturated vapour Hvap = 2779.7 J/g

Enthalpy of steam at 101.33 kPa and 105°C is H2= 2686.1 J/g

Step 3.

In constant enthalpy process, H1=H2 which means inlet enthalpy is equal to outlet enthalpy

So, H1=H2

H2= (1-x)Haq+XHvap.........1

Putting the values in 1

2686.1(J/g) = {(1-x)x 781.124(J/g)} + {X x 2779.7 (J/g)}

= 781.124 (J/g) - x781.124 (J/g) = x2779.7 (J/g)

1904.976 (J/g) = x1998.576 (J/g)

x = 1904.976 (J/g)/1998.576 (J/g)

x = 0.953

So, the quality of the wet steam is 0.953

7 0

3 years ago

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