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vampirchik [111]
1 year ago
8

At 850K, 65L of gas has a pressure of 450kPa. What is the volume (in liters) if the gas is cooled to 430K and the pressure decre

ases to 325kPa?
Chemistry
1 answer:
Novosadov [1.4K]1 year ago
8 0

Answer:

V₂ =  45.53 L

Explanation:

Given data:

Initial temperature = 850 K

Initial volume = 65 L

Initial pressure = 450 KPa

Final temperature = 430 K

Final pressure = 325 KPa

Final volume = ?

Solution:

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

V₂ = P₁V₁ T₂/ T₁ P₂  

V₂ = 450 KPa× 65 L × 430 K / 850 K × 325KPa  

V₂ = 12577500 KPa .L. K / 276250 K. KPa

V₂ =  45.53 L

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Answer:

It sounds like they are studying French phonemes

Explanations:

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7 0
1 year ago
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If a swimming pool contains 2,850 kiloliters (kL) of water, how many gallons (gal) of water does it contain? (1 gal = 3.785 L)
siniylev [52]

we are given

a swimming pool contains 2,850 kiloliters (kL) of water

2,850 kiloliters (kL)=2850000L

we know that

1gal=3.785L

3.785L=1gal

Firstly , we will find for 1L

1L=\frac{1}{3.785}gal

now, we can multiply both sides by 2850000

2850000*1L=2850000*\frac{1}{3.785}gal

2850000L=2850000*\frac{1}{3.785}gal

2850000L=752972.25892gal

so,

a swimming pool contains 752972.25892gal of water...........Answer


6 0
2 years ago
Read 2 more answers
Two students titrated a 25.0 mL aliquot of pear juice with 0.107 M NaOH to the phenolphthalein end point. The initial buret read
Olin [163]

Answer:

0.363g citric acid

Explanation:

Sodium hydroxide (NaOH) reacts with acids, thus:

NaOH + H⁺ → H₂O + Na⁺

The volume of titration is:

18.39mL - 0.73mL = 17.66mL

Moles of this volume in 0.107M NaOH are:

0.01766L × (0.107 mol / L) = 0.00189mol NaOH ≡ mol citric acid<em> -Assuming the only acid in pear juice is citric acid-</em>

As molar mass of citric acid is 192.124g/mol:, Mass of citric acid is:

0.00189mol citric acid × (192.124g / mol) = <em>0.363g citric acid</em>

I hope it helps!

6 0
2 years ago
A 2.50 g sample of zinc is heated, and then placed in a calorimeter containing 65.0 g of water. Temperature of water increases f
swat32

718.65 degrees is the initial temperature of the zinc metal sample.

Explanation:

Data given:

mass of zinc sample = 2.50 grams

mass of water  = 65 grams

initial temperature of water = 20 degrees

final temperature of water = 22.5 degrees

ΔT  = change in temperature of water is 2.50 degrees

specific heat capacity of zinc cp= 0.390 J/g°C

initial temperature of zinc sample = ?

cp of water = 4.186 J/g°C

heat absorbed = heat released (no heat loss)

formula used is

q = mcΔT

q water = 65 x 4.286 x 2.5

q water = 696.15 J

q zinc = 2.50 x 0.390 x (22.50- Ti)

equating the two equations

696.15 = - 22.50+ Ti

Ti = 718.65 degrees is the initial temperature of zinc.

6 0
2 years ago
Compound X has a molar mass of 104.01·gmol−1 and the following composition: element mass % nitrogen 26.93% fluorine 73.07% Write
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Answer:

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