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Setler [38]
3 years ago
15

One balloon was only partly filled when put on the plane. During the flight, the balloon volume

Chemistry
1 answer:
iren2701 [21]3 years ago
7 0

Answer:

P₂ = 0.8 atm

Explanation:

Given data:

Initial pressure = 1 atm

Initial volume = 7 L

Final volume = 9 L

Final pressure = ?

Solution:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = Initial volume

P₂ = Final pressure

V₂ = Final volume

Now we will put the vales in formula.

P₁V₁ = P₂V₂

P₂ = P₁V₁ /V₂

P₂ = 1 atm × 7 L / 9 L

P₂ = 7 atm. L / 9 L

P₂ = 0.8 atm

You might be interested in
How many centimeters are in .479 kilometers
trapecia [35]

Answer:

47900 cm

General Formulas and Concepts:

<u>Chemistry</u>

  • Base 10 Decimal System
  • Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

0.479 km

<u>Step 2: Identify Conversions</u>

1 km = 1000 m

1 m = 100 cm

<u>Step 3: Convert</u>

<u />0.479 \ km(\frac{1000 \ m}{1 \ km} )(\frac{100 \ cm}{1 \ m} ) = 47900. cm

<u>Step 4: Check</u>

<em>We are given 3 sig figs. Follow sig fig rules.</em>

47900. cm ≈ 47900 cm

7 0
3 years ago
All charges in a compound must have ?
MakcuM [25]
They must have both cations and anions.

I hope this helps.
8 0
4 years ago
The Ksp of calcium carbonate in water at 25 °C is 2.25 x 10-8. CaCO3(s) &lt;----&gt; Ca2+ (aq) + CO3 2- (aq) What is favored at
seraphim [82]

Answer:

the solubility of CaCO3 is 0.015g/l 25 °C

is favored at equilibrium

Explanation:

The Ksp of calcium carbonate in water at 25 °C is 2.25 x 10-8. CaCO3(s) <----> Ca2+ (aq) + CO3 2- (aq) What is favored at equilibrium?

solubility is the property of a solute to dissolve in a solvent(liquid, gas ) to form a solution(soution can be saturated ,unsaturated, or supersaturated)

CaCO3(s) <----> Ca2+ (aq) + CO3 2- (aq)

in partial dissociation , we can say

2.25x 10^-8=Ca^{2+} +CO^{2-}_{3} }

let Ca^2+=CO3^-2=S

2.25x10^-8=S*S

S^2=2.25x10^-8

S=0.00015mol/L

Converting that to g/l

the relative molecular mass of CaCO3=100g/mol

0.00015*100g/mol

0.015g/l

the solubility of CaCO3 is 0.015g/l @room temperature

is favored   at equilibrium

5 0
3 years ago
How many calories of heat are necessary to raise the temperature of 319.5 g of water from 35.7 °C
rjkz [21]

20600Cal              

Explanation:

Given parameters:

Mass of water = 319.5g

Initial temperature = 35.7°C

Final temperature = 100°C

Unknown:

Calories needed to heat the water = ?

Solution:

The calories is the amount of heat added to the water. This can be determined using;

     H  =   m  c Ф

c  = specific heat capacity of water = 4.186J/g°C

   H is the amount of heat

    Ф is the change in temperature

    H = m c (Ф₂ - Ф₁)

    H = 319.5 x 4.186 x (100 - 35.7) = 85996.56J

Now;

     1kilocalorie = 4184J

     

85996.56J to kCal; \frac{85996.56}{4184}   = 20.6kCal  = 20600Cal

               

learn more:

Specific heat brainly.com/question/3032746

#learnwithBrainly

6 0
3 years ago
Which way of dissolving changes the chemical identity of the solute?
sasho [114]

Answer:

Ionization and dissociation

5 0
3 years ago
Read 2 more answers
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