All categories

3 years ago

What is the volume of 11.2 g of O2 at 7.78 atm and 415 K?

Chemistry

1 answer:

irina1246 [14]3 years ago

3 0

Answer:

1.53 L

Explanation:

Step 1: Given data

Mass of oxygen (m): 11.2 g

Pressure (P): 7.78 atm

Temperature (T): 415 K

Ideal gas constant (R): 0.0821 atm.L/mol.K

Step 2: Calculate the moles (n) corresponding to 11.2 g of oxygen

The molar mass of oxygen is 32.00 g/mol.

11.2 g × (1 mol/32.00 g) = 0.350 mol

Step 3: Calculate the volume of oxygen

We will use the ideal gas equation.

P × V = n × R × T

V = n × R × T / P

V = 0.350 mol × (0.0821 atm.L/mol.K) × 415 K / 7.78 atm

V = 1.53 L

You might be interested in

The digestion process starts in the _____.<br> esophagus<br> stomach<br> small intestine<br> mouth

ahrayia [7]

Your mouth is where the digestion process starts because when yo chew your food, nutrients is released into your body

7 0

2 years ago

Read 2 more answers

A speed-time graph shows a line sloping downward. How was the speed changing?

Mice21 [21]

It means it has a constant deceleration

3 0

2 years ago

The radioactive isotope of lead, Pb-209, decays at a rate proportional to the amount present at time t and has a half-life of 3.

Marina86 [1]

Answer:

Therefore it will take 7.66 hours for 80% of the lead decay.

Explanation:

The differential equation for decay is

$\frac{dA}{dt}= kA$

$\Rightarrow \frac{dA}{A}=kdt$

Integrating both sides

ln A= kt+c₁

$\Rightarrow A= e^{kt+c_1}$

$\Rightarrow A=Ce^{kt}$ [ $e^{c_1}=C$ ]

The initial condition is A(0)= A₀,

$\therefore A_0=Ce^{0.k}$

$\Rightarrow C=A_0$

$\therefore A=A_0e^{kt}$ .........(1)

Given that the $Pb_{209}$ has half life of 3.3 hours.

For half life $A=\frac12 A_0$ putting this in equation (1)

$\frac12A_0=A_0e^{k\times3.3}$

$\Rightarrow ln(\frac12)= 3.3k$ [taking ln both sides, $ln \ e^a=a$ ]

$\Rightarrow k=\frac{ln \frac12}{3.3}$

⇒k= - 0.21

Now A₀= 1 gram, 80%=0.8

and A= (1-0.8)A₀ = (0.2×1) gram = 0.2 gram

Now putting the value of k,A and A₀in the equation (1)

$\therefore 0.2=1e^{(-0.21)\times t}$

$\Rightarrow e^{-0.21t}=0.2$

$\Rightarrow -0.21t= ln(0. 2)$

$\Rightarrow t= \frac{ln (0.2)}{-0.21}$

⇒ t≈7.66

Therefore it will take 7.66 hours for 80% of the lead decay.

3 0

2 years ago

Which of the following would have the largest pKa?

garri49 [273]

Answer:

CH3CH2NH3+/CH3CH2NH2 would have the largest pKa

Explanation:

To answer this question we must know Kb of CH3CH2NH2 is 5.6x10⁻⁴, and for C6H5NH2 is 4.0x10⁻¹⁰. And the CH3CH2NH3+ and C6H5NH3+ are related with these substances because are their conjugate base. That means:

pKa of CH3CH2NH3+ = CH3CH2NH2; C6H5NH3+ = C6H5NH2

Also, Kw / Kb = Ka

Thus:

pKa of CH3CH2NH3+/CH3CH2NH2 is:

Kw / kb = Ka = 1.79x10⁻¹¹

-log Ka = pKa

pKa = 10.75

pKa of C6H5NH3+/ C6H5NH2 is:

Kw / kb = Ka = 2.5x10⁻⁵

-log Ka = pKa

pKa = 4.6

That means CH3CH2NH3+/CH3CH2NH2 would have the largest pKa

5 0

3 years ago

Asap Help!!! I just wanted to know if my answer is correct i am not sure if its not please correct me

Elenna [48]

Answer:

Precipitate

Explanation:

A precipitate is a solid formed from a chemical solution

7 0

1 year ago

Read 2 more answers