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2 years ago

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A 2 L bottle containing only air is sealed at a temperature of 22°C and a pressure of 0.982 atm. The bottle is placed in a freez

er and allowed to cool to -3°C. What is the pressure in the bottle if the volume changes to 1.8 L?

1 answer:

Degger [83]2 years ago

5 0

Answer:

.997 atm

Explanation:

1. Find the combined gas law formula...

(P1V1/T1 = P2V2/T2)

2. Find our numbers...

P1= .982 atm

P2= ? (trying to find)

V1= 2 L

V2= 1.8 L

T1= 22 C = 295 K

T2= -3 C = 270 K

- Note: always use Kelvin. To find Kelving add 273 to ___C.

3. Rearrange formula to fit problem...

(P2=P1V1T2/V2T1)

4. Fill in our values...

P2= .982 atm x 2 L x 270 K / 1.8 L x 295 K

5. Do the math and your answer should be...

.997 atm

- If you need more help or still do not understand please let me know and I would be glad to help!

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A student dissolved 4.00 g of Co(NO3)2 in enough water to make 100. mL of stock solution. He took 4.00 mL of the stock solution

mihalych1998 [28]

Answer:

0.08097 grams of nitrate ions are there in the final solution.

Explanation:

Moles of cobalt(II) nitrate ,n= $\frac{4.00 g}{245 g/mol}=0.01633 mol$

Volume of the cobalt(II) nitrate solution, V = 100.0 mL = 0.1 L

$Molarity=\frac{n}{V(L)}$

Let the molarity of the solution be $M_1$

$M_1=\frac{0.01633 mol}{0.1 L}=0.1633 M$

A students then takes 4 .00 mL of $M_1$ solution and dilute it to 275 ml.

$M_1=0.1633 M$

$V_1=4.00 mL$

$M_2=?$ (molarity after dilution)

$V_2=275 mL$ (after dilution)

$M_1V1=M-2V_2$

$M_2=\frac{M_1V_1}{V_2}=\frac{0.1633 M\times 4.00 mL}{275 mL}=0.002375 M$

Molarity of the of solution after dilution is 0.002375 M.

$Co(NO_3)_2(aq)\rightarrow Co^{2+}(aq)+2NO_3^{-}(aq)$

1 mol of cobalt(II) nitrate gives 2 moles of nitrate ions. Then 0.002375 M solution of cobalt (II) nitrate will give:

$[NO_3^{-}]=\frac{2}{1}\times 0.002375 M=0.004750 M$

Moles of nitrate ions = n

Volume of the solution = 275 mL = 0.275 L

Molarity of the nitrate ions = $[NO_3^{-}]=0.004750 M$

$[NO_3^{-}]=\frac{n}{0.275 L}$

n = 0.001306 mol

Mass of 0.001306 moles of nitrate ions:

0.001306 mol × 62 g/mol= 0.08097 g

0.08097 grams of nitrate ions are there in the final solution.

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2 years ago

Carbon dating of small bits of charcoal used in cave paintings has determined that some of the paintings are from 10000 to 30000

Andru [333]

The age of painting was determined from the decay kinetics of the radioactive Carbon -14 present in the painting sample.

Given that the half life of Carbon-14 is 5730 years.

Radioactive decay reactions follow first order rate kinetics.

Calculating the decay constant from half life:

λ $= \frac{0.693}{t_{1/2} }$

= $\frac{0.693}{5730 yr}$ = $1.21*10^{-4}yr^{-1}$

Setting up the radioactive rate equation:

$ln\frac{A_{t} }{A_{0} } =-kt$

Where $A_{t} = Activity after time t = 0.80microCi$

$A_{t} = initial activity = 6.4microCi$

k = decay constant = $1.21*10^{-4}yr^{-1}$

$ln\frac{0.80uCi}{6.4uCi} =-(1.21*10^{-4}yr^{-1})t$

ln 0.125 = $-(1.21*10^{-4}yr^{-1})t$

-2.079= $-(1.21*10^{-4}yr^{-1})t$

t= $\frac{2.07944}{1.21*10^{-4} } yr$

= 17185 years

t = 17185 years

Therefore age of the painting based in the radiocarbon -14 dating studies is 17185 years

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2 years ago

5. Gabi has plans with her friends to go to a concert on her birthday in 4 days. She is so excited that she wants to know how ma

drek231 [11]

Answer:

So she is very anxious because she has to wait 345600 seconds

Explanation:

60 second = 1 minute

60 minute = 1 hour

1 hour has 3600 seconds (60*60)

24 hour = 1 day

3600 second * 24 hours =

1 day has 86400 seconds so in four days

86400 * 4 = 345600

7 0

2 years ago

What is the reaction energy Q of this reaction? Use c2=931.5MeV/u. Express your answer in millions of electron volts to three si

emmasim [6.3K]

Answer:

Energy= 2.7758 × 10^-11 J ;

71.112×10^-6 kJ.

Mass defect in Kilogram= 3.0885×10^-28 kg.

That is; 3.1×10^-28 kg(to two significant figure).

Explanation:

(Note: Check equation of reaction in the attached file/picture).

STEP ONE: we have to calculate the Mass defect.

Mass defect= Mass of reactants -- Mass of products.

Mass of the products: (140.9144+91.9262+3.060) u.

= 235.8666 u.

Mass of reactants: (1.0087+235.0439) u= 236.0526 u.

Therefore, the Mass defect= (236.0526 -- 235.8666) u

= 0.1860 u.

STEP TWO: Converting the Mass defect to energy;

0.18860 × 1.6605 × 10^27 kg

= 3.0885× 10^-28 kg

STEP THREE: Calculating energy released . Recall(from the question) c^2= 931.5 Mev/u. This is also equals to 9×10^16 m/s.

E=Mc^2.

Where E= energy released, c= speed of light, M= Mass.

Slotting in the values;

E= 3.0885×10^-28 kg × 9×10^16 m/s.

E=2.7758 × 10^-11 J.

Know that;( 1g of uranium × 1 mol of uranium ÷ 235.0439 g of uranium) × (6.002×10^23 atom of uranium/ 1 mol of uranium) × 2.7758× 10^-11.

=7.1112×10^-10 J

= 71.112×10^6 kJ.

3 0

2 years ago

What is the oxidation state of selenium in SeO3?

ra1l [238]

Answer: The oxidation state of selenium in SeO3 is +6

Explanation:

SeO3 is the chemical formula for selenium trioxide.

- The oxidation state of SeO3 = 0 (since it is stable and with no charge)

- the oxidation number of oxygen (O) IN SeO3 is -2

- the oxidation state of selenium in SeO3 = Z (let unknown value be Z)

Hence, SeO3 = 0

Z + (-2 x 3) = 0

Z + (-6) = 0

Z - 6 = 0

Z = 0 + 6

Z = +6

Thus, the oxidation state of selenium in SeO3 is +6

8 0

2 years ago