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2 years ago

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What will the pressure be if 89.9 moles of argon are contained in a 12.0 L cylinder that is pressurized at a temperature of 300

K?

1 answer:

irinina [24]2 years ago

7 0

P=nRT/V
p=89.9(8.314)(12)/300
P=8969.14/300
P=29.89atm
P=29.9atm

Done!

You might be interested in

Indicate whether the following are acidic, basic, or neutral solution

goldenfox [79]

You answer this by using the pH formula and and the relation of pH and pOH, pH = -log[H+] and 14 = pH + pOH. The correct classification are as follows:

<span>A. [H2O+]=6.0x10^-12
basic

B. [H3O+]=1.4x10^-9
basic

C. [OH-]=5.0x10^-12
acidic

D. {OH-]=3.5x10^-10
acidic

Hope this answers the question.

</span>

5 0

3 years ago

Study this chemical reaction: (aq)(s)(s)(aq) Then, write balanced half-reactions describing the oxidation and reduction that hap

disa [49]

The question is incomplete, complete question is:

Study this chemical reaction:

$FeSO_4 (aq) + Zn (s)\rightarrow Fe (s) + ZnSO_4 (aq)$

Then, write balanced half-reactions describing the oxidation and reduction that happen in this reaction.

Oxidation:

Reduction:

Answer:

Oxidation taking place in given reaction :

$Zn(s)\rightarrow Zn^{2+}+2e^-$

Reduction taking place in given reaction;

$Fe^{2+}(aq)+2e^-\rightarrow Fe(s)$

Explanation:

Redox reaction is defined as the reaction in which oxidation and reduction reaction occur side by side.

Oxidation reaction is defined as the chemical reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.

$X\rightarrow X^{n+}+ne^-$

Reduction reaction is defined as the chemical reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.

$X^{n+}+ne^-\rightarrow X$

$FeSO_4 (aq) + Zn (s) \rightarrow Fe (s) + ZnSO_4 (aq)$

In the given reaction, iron(II) ions are getting reduced and zinc metal is getting oxidized to zinc(II) ions.

Oxidation :

$Zn(s)\rightarrow Zn^{2+}+2e^-$

Reduction ;

$Fe^{2+}(aq)+2e^-\rightarrow Fe(s)$

6 0

3 years ago

Polonium is a rare element with 33 radioisotopes. The most common one, 210Po, has 84 protons and 126 neutrons. When 210Po decays

Fofino [41]

Answer:

Lead Pb

Explanation:

Firstly, we need to know what occurs when a radioisotope emits an alpha particle. An alpha particle is an helium atom. When an isotope emits an alpha particle, it loses an helium atom corresponding to subtracting 4 from its mass number and 2 from its atomic number. This of course coupled with the release of radiation.

Now, we polonium has a proton number of 84 and a mass number of 210. Subtracting 2 and 4 respectively from its proton and mass numbers will yield 82 and 206 proton and mass numbers respectively.

Hence, the decomposition of the Po-210 isotope will yield an element with 82 proton number and 206 mass number. This corresponds to the element Lead.

210Po ——> 206Pb + alpha particle + radiation

6 0

3 years ago

What pressure is exerted by 0.625 moles of a gas in a 45.4 L container at -24.0 °C? Find pressure in units of atm.

scoundrel [369]

Answer:

Ok to solve this you will need to use the Ideal Gas Law Formula which is as follows:

PV = nRT

P= pressure

V= volume

n= # of moles

R= Universal Gas Constant (0.0821 L x atm/mol x K)

T= Kelvin temperature

1.Simplify the Ideal Gas Law formula to what you need to solve for:

P = (nRT)/ V

2. List all you components as follows (this makes the process easier):

P = ?

V = 45.4 L

n = 0.625 mol

R = 0.0821 L x atm/ mol x K

T = 249 K

To find the Kelvin temperature K = C + 273

3. Plug in all your components in your set up formula:

P = [(0.625 mol)(0.0821 L x atm/ mol x K)(249 K)] / (45.4 L)

4. Cross out all similar units so the only thing left is atm because you are trying to find pressure.

P = [(0.625)(0.0821atm)(249)] / (45.4)

5. Multiply through and simplify

P = 0.28 atm

B. is the correct answer.

Glad I could help!! If you have any other questions just message me. Hopefully this was helpful.

Explanation:

3 0

3 years ago

A chemist needs to determine the concentration of a sulfuric acid solution by titration with a standard sodium hydroxide solutio

brilliants [131]

Explanation:

The given data is as follows.

Concentration of standard NaOH solution = 0.1922 M

Let the original acid solution concentration be x.

$M_{1}V_{1} = M_{2}V_{2}$

$x \times 25 = 250 \times M_{2}$

$M_{2} = \frac{x \times 25}{250}$

= 0.1 x M

$V_{2}$ = 10.00 mL (given)

The reaction equation is as follows.

$2NaOH + H_{2}SO_{4} \rightarrow Na_{2}SO_{4} + 2H_{2}O$

Concentration × Volume of $H_{2}SO_{4}$ = Concentration × Volume of NaOH

$\frac{M_{2}V_{2}}{n_{2}} = \frac{M_{3}V_{3}}{n_{3}}$

$\frac{0.1 x \times 10 ml}{1} = \frac{0.1922 M \times 13.68 ml}{2}$

x = 1.314 M

Therefore, we can conclude that the concentration of the original sulfuric acid solution is 1.314 M.

7 0

3 years ago