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3 years ago

15

2. If a gas at 500 mL has a temperature of 45°C, then what is the new volume when the temperature is increased to 65°C? Show you

r work.

1 answer:

OLEGan [10]3 years ago

8 0

Answer:

V2= 531.4 ml

Explanation:

T1= 273+45 °C= 318 Kelvin

T2= 273+65 °C= 338 Kelvin

V1= 500ml *( 0.001L/1ml)= 0.5 Liters

V2= $\frac{V1 * T2}{T1}$ = $\frac{0.5 L * 338K}{318K}$ = 0.5314465409 Liters* (1 ml/0.001 L)= 531.4 ml

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Gaseous butane reacts with gaseous oxygen gas to produce gaseous carbon dioxide and gaseous water . If of water is produced from

krek1111 [17]

The given question is incomplete. The complete question is :

Gaseous butane reacts with gaseous oxygen gas to produce gaseous carbon dioxide and gaseous water . If 1.31g of water is produced from the reaction of 4.65g of butane and 10.8g of oxygen gas, calculate the percent yield of water. Be sure your answer has the correct number of significant digits in it.

Answer: 28.0 %

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of butane}=\frac{4.65g}{58g/mol}=0.080moles$

$\text{Moles of oxygen}=\frac{10.8g}{32g/mol}=0.34moles$

$2C_4H_{10}(g)+13O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)$

According to stoichiometry :

13 moles of $O_2$ require 2 moles of butane

Thus 0.34 moles of $O_2$ will require= $\frac{2}{13}\times 0.34=0.052moles$ of butane

Thus $O_2$ is the limiting reagent as it limits the formation of product and butane is the excess reagent.

As 13 moles of $O_2$ give = 10 moles of $H_2O$

Thus 0.34 moles of $O_2$ give = $\frac{10}{13}\times 0.34=0.26moles$ of $H_2O$

Mass of $H_2O=moles\times {\text {Molar mass}}=0.26moles\times 18g/mol=4.68g$

${\text {percentage yield}}=\frac{\text {Experimental yield}}{\text {Theoretical yield}}\times 100\%$

${\text {percentage yield}}=\frac{1.31g}{4.68g}\times 100\%=28.0\%$

The percent yield of water is 28.0 %

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4 years ago

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sdas [7]

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3 years ago

Calculate the wavelength of the photon emitted when an electron makes a transition from n=6 to n=3. You can make use of the foll

Angelina_Jolie [31]

<u>Answer:</u> The wavelength of light is $1.094\times 10^{-6}m$

<u>Explanation:</u>

To calculate the wavelength of light, we use Rydberg's Equation:

$\frac{1}{\lambda}=R_H\left(\frac{1}{n_f^2}-\frac{1}{n_i^2} \right )$

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant = $1.097\times 10^7m^{-1}$

$n_f$ = Final energy level = 3

$n_i$ = Initial energy level = 6

Putting the values in above equation, we get:

$\frac{1}{\lambda }=1.097\times 10^7m^{-1}\left(\frac{1}{3^2}-\frac{1}{6^2} \right )\\\\\lambda =\frac{1}{914617m^{-1}}=1.094\times 10^{-6}m$

Hence, the wavelength of light is $1.094\times 10^{-6}m$

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4 years ago