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Sindrei [870]
3 years ago
13

a balloon has a pressure of 104.5 kpa and a volume of 3.4L. If the pressure is reduced to 42.7 kpa, what will the volume change

to?​
Chemistry
2 answers:
Nuetrik [128]3 years ago
7 0

Answer:

The answer to your question is 8,32 L

Explanation:

Data

Pressure 1 = 104.5 kPa

Volume 1 = 3.4 L

Pressure 2 = 42.7 kPa

Volume 2 = ?

Formula

- To solve this problem use the Boyle's law.

       P1V1 = P2V2

- Solve for V2

       V2 = P1V1 / P2

- Substitution

       V2 = (104.5)(3.4) / 42.7

-Simplification

       V2 = 355.3 / 42.7

- Result

        V2 = 8.32 L            

Nookie1986 [14]3 years ago
3 0

Answer:

The new volume is 8.3 L

Explanation:

Step 1: Data given

Initial pressure of balloon = 104.5 kPa = 1.0313348 atm

Volume of the balloon is 3.4 L

The pressure is reduced to 42.7 kPa = 0.4214162 atm

Step 2: Calculate the new volume

p1*V1 = p2 *V2

⇒p1 = the initial pressure in the balloon = 1.0313348 atm

⇒V1 = the initial volume = 3.4 L

⇒p2 = the reduced pressure = 0.4214162 atm

⇒V2 = the new volume

1.0313348 atm * 3.4 L = 0.4214162 * V2

V2 = 8.3 L

The new volume is 8.3 L

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How many atoms are there in 4430 grams of lithium (Li)?
Ronch [10]

Answer:

3843.96 × 10²³ atoms

Explanation:

Given data:

Mass of Li = 4430 g

Number of atoms = ?

Solution:

We will calculate the number of moles first.

Number of moles =mass/molar mass

Number of moles = 4430 g/ 6.94 g/mol

Number of moles = 638.32 mol

1 mole contain 6.022 × 10²³ atoms

638.32 mol × 6.022 × 10²³ atoms / 1 mol

3843.96 × 10²³ atoms

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podryga [215]

Answer :

Part 1 : Balanced reaction, 3H_2(g)+N_2(g)\rightarrow 2NH_3(g)

Part 2 : The theoretical yield of NH_3 gas = 440.96 g

Part 3 : The % yield of ammonia is 90.03 %

Solution : Given,

Mass of N_2 = 475 g

Molar mass of N_2 = 28 g/mole

Molar mass of NH_3 = 17 g/mole

Experimental yield of NH_3 = 397 g

<u>Answer for Part (1) :</u>

The balanced chemical reaction is,

3H_2(g)+N_2(g)\rightarrow 2NH_3(g)

<u>Answer for Part (2) :</u>

First we have to calculate the moles of N_2.

\text{Moles of }N_2=\frac{\text{ Mass of }N_2}{\text{ Molecular mass of }N_2}=\frac{475g}{28g/mole}=16.96moles

From the given reaction, we conclude that

1 moles of N_2 gas react to give 2 moles of NH_3 gas

16.96 moles of N_2 gas react to give \frac{2}{1}\times 16.96=33.92 moles of NH_3 gas

Now we have to calculate the mass of NH_3 gas.

\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3

\text{ Mass of }NH_3=(33.92moles)\times (17g/mole)=440.96g

Therefore, the theoretical yield of NH_3 gas = 440.96 g

<u>Answer for Part (3) :</u>

Formula used for percent yield :

\% \text{ yield of }NH_3=\frac{\text{ Experimental yield of }NH_3}{\text{ Theoretical yield of }NH_3}\times 100

\% \text{ yield of }NH_3=\frac{397g}{440.96g}\times 100=90.03\%

Therefore, the % yield of ammonia is 90.03 %

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Explanation:

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The molecular weight of NaCl is 58.44 grams/mole. If you had a 1.0 molar solution (1.0 M), you would have to put 58.44 g of salt
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Answer:

5,844 grams of NaCl

Explanation:

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58,44 x 0,1 = 5,844 grams of NaCl

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