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3 years ago

Of the following two gases,

Chemistry

1 answer:

kap26 [50]3 years ago

4 0

Answer:

I think C12 will diffuse faster as it is less dense than C02.

Explanation:

I hope this helps you.

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1. A sample of gas has an initial volume of 25 L and an initial pressure of 123.5 kPa. If the pressure changes to

sveta [45]

Answer:

$\large \boxed{\text{17.mL}}$

Explanation:

The temperature and amount of gas are constant, so we can use Boyle’s Law.

$p_{1}V_{1} = p_{2}V_{2}$

Data:

$\begin{array}{rcrrcl}p_{1}& =& \text{123.5 kPa}\qquad & V_{1} &= & \text{25 L} \\p_{2}& =& \text{179.9 kPa}\qquad & V_{2} &= & ?\\\end{array}$

Calculations:

$\begin{array}{rcl}123.5 \times 25 & =& 179.9V_{2}\\3088 & = & 179.9V_{2}\\V_{2} & = &\dfrac{3088}{179.9}\\\\& = &\textbf{17 L}\\\end{array}\\\text{The new volume of the gas is } \large \boxed{\textbf{17 L}}$

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3 years ago

What human life process requires hydrochloric acid? <br><br> THANKS!!!

notka56 [123]

The answer is Enzyme

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3 years ago

Research the compositions of Pennies. What was the composition of each of your Pennies prior to treatment

LenKa [72]

Answer:

History of composition

Years Material Weight (grains)

1944–1946 gilding metal (95% copper, 5% zinc) 48 grains

1947–1962 bronze (95% copper, 5% tin and zinc) 48 grains

1962 – September 1982 gilding metal (95% copper, 5% zinc) 48 grains

October 1982 – present copper-plated zinc (97.5% zinc, 2.5% copper) 38.6 grains

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2 years ago

Force acts on it.

BartSMP [9]

Answer:D.

Explanation:

4 0

3 years ago

Kc for the reaction N2O4 <=> 2NO2 is 0.619 at 45 degrees C If 50.0g of N2O4 is introduced into an empty 2.10L container, w

Nadya [2.5K]

Answer:

p(N2O4) = 0.318 atm

p(NO2) = 7.17 atm

Explanation:

Step 1: Data given

Kc = 0.619

Temperature = 45.0 °C

Mass of N2O4 = 50.0 grams

Volume = 2.10 L

Molar mass N2O4 = 92.01 g/mol

Step 2: The balanced equation

N2O4 ⇔ 2NO2

Step 3: Calculate moles N2O4

Moles N2O4 = 50.0 grams / 92.01 g/mol

Moles N2O4 = 0.543 moles

Step 4: The initial concentration

[N2O4] = 0.543 moles/2.10 L = 0.259 M

[NO2]= 0 M

Step 5: Calculate concentration at the equilibrium

For 1 mol N2O4 we'll have 2 moles NO2

[N2O4] = (0.259 -x)M

[NO2]= 2x

Step 6: Calculate Kc

Kc = 0.619= [NO2]² / [N2O4]

0.619 = (2x)² / (0.259-x)

0.619 = 4x² / (0.259 -x)

x = 0.1373

Step 7: Calculate concentrations

[N2O4] = (0.259 -x)M = 0.1217 M

[NO2]= 2x = 0.2746 M

Step 8: The moles

Moles = molarity * volume

Moles N2O4 = 0.1217 M * 2.10 = 0.0256 moles

Moles NO2 = 0.2746 M * 2.10 = 0.577 moles

Step 9: Calculate partial pressure

p*V = n*R*T

⇒ with p = the partial pressure

⇒ with V = the volume = 2.10 L

⇒ with n = the number of moles

⇒ with R = the gas constant = 0.08206 L*atm/mol*K

⇒ with T = the temperature = 45 °C = 318 K

p = (nRT)/V

p(N2O4) = (0.0256 *0.08206 * 318)/ 2.10

p(N2O4) = 0.318 atm

p(NO2) = (0.577 *0.08206 * 318)/ 2.10

p(NO2) = 7.17 atm

6 0

3 years ago