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Alexxandr [17]
2 years ago
6

In Vancouver where the air pressure is 98kPa, a child is given a 2.7 L ballon. The family then drives to Banff which is 1386 m a

bove sea level and the air pressure is only 85 kPa. If the temperature is the same in both places, what is the volume of the baloon in Banff?.
Chemistry
1 answer:
iVinArrow [24]2 years ago
5 0

The volume of the balloon in Banff is 3.113 liters.

According to the equation of state for ideal gases, the pressure (P), in kilopascals, is inversely proportional to the volume (V), in liters. The <em>final</em> volume of the balloon is found by the following relationship:

V_{2} = \frac{P_{1}\cdot V_{1}}{P_{2}} (1)

Where:

  • V_{1} - Initial volume, in liters.
  • V_{2} - Final volume, in liters.
  • P_{1} - Initial pressure, in kilopascals.
  • P_{2} - Final pressure, in kilopascals.

If we know that P_{1} = 98\,kPa, P_{2} = 85\,kPa and V_{1} = 2.7\,L, then the final volume of the balloon is:

V_{2} = 2.7\,L\times \frac{98\,kPa}{85\,kPa}

V_{2} = 3.113\,L

The volume of the balloon in Banff is 3.113 liters.

We kindly invite to check this question on ideal gases: brainly.com/question/16211117

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anzhelika [568]

1.2 x 10^10 mm^3 is equivalent to 12.0 m^3

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2 years ago
!: (a) 20 cm: 80
RoseWind [281]

Answer:

THE LENGTH OF THE AIR COLUMN IS 9.5 CM

Explanation:

Taking the atmospheric pressure to be 760 mmHg;

When the capillary tube is held horizontally, the pressure of the tube is 760 mmHg

when the capillary tube is held vertically, the pressure increases by 4 cm = 40 mm

The new pressure of the tube is hence, 760 + 40 mmHg = 800 mmHg

Using the pressure forlmula;

P1 V1 = P2 V2

P1 A1 L1 = P2 A2 L2

where A1 and A2 is the area of the capillary tube and it is equal, it cancels out.

P1 l1 = P2 l2

l2 = P1 l1 / P2

l2 = 760 * 10 / 800

l2 = 9.5 cm

The length of the air in the tube is 9.5 cm.

5 0
3 years ago
How many grams are in 44.8 liters of nitrogen gas, n2?<br> a) 56g<br> b) 27g<br> c) 36g<br> d) 112g
kkurt [141]

a) 56g

<h3>Calculation:</h3>

At STP,

22.4 L of N₂ = 1 mol

We have given 44.8 L of N₂, therefore,

44.8 L of N₂ = \frac{44.8}{22.4}

                    = 2 mol

We know that,

1 mol of N₂ = 28 g

Hence,

2 mol of N₂ = 28 × 2

                   = 56g

Hence, there are 56 g of N₂ in 44.8 L of nitrogen gas.

Learn more about calculation at STP here:

brainly.com/question/9509278

#SPJ4

3 0
2 years ago
Read 2 more answers
A 0.708 g sample of a metal, M, reacts completely with sulfuric acid according to the reaction M ( s ) + H 2 SO 4 ( aq ) ⟶ MSO 4
ollegr [7]

Answer:

The metal has a molar mass of 65.37 g/mol

Explanation:

Step 1: Data given

Mass of the metal = 0.708 grams

Volume of hydrogen = 275 mL = 0.275 L

Atmospheric pressure = 1.0079 bar = 0.9947 atm

Temperature = 25°C

Vapor pressure of water at 25 °C = 0.03167 bar = 0.03126 atm

Step 2: The balanced equation

M(s) + H2SO4(aq) ⟶ MSO4 (aq) + H2(g)

Step 3: Calculate pH2

Atmospheric pressure = vapor pressure of water + pressure of H2

0.9947 atm = 0.03126 atm + pressure of H2

Pressure of H2 = 0.9947 - 0.03126

Pressure of H2 = 0.96344 atm

Step 4: Calculate moles of H2

p*V=n*R*T

⇒ with p = The pressure of H2 = 0.96344 atm

⇒ with V = the volume of H2 = 0.275 L

⇒ with n = the number of moles H2 = TO BE DETERMINED

⇒ with R = the gas constant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 25°C = 298 Kelvin

n = (p*V)/(R*T)

n = (0.96344 * 0.275)/(0.08206*298)

n = 0.01083 moles

Step 5: Calculate moles of M

For 1 mole of H2 produced, we need 1 mole M

For 0.0108 moles of H2 we need 0.01083 moles of M

Step 6: Calculate molar mass of M

Molar mass M = Mass M / moles M

Molar mass M = 0.708 grams / 0.01083 moles

Molar mass M = 65.37 g/mol

The metal has a molar mass of 65.37 g/mol

5 0
3 years ago
Find the mass, in grams, of 3.758 mol CH4. ( show work )
vladimir2022 [97]

Answer:

The mass of CH4 is 60, 29 grams.

Explanation:

We use the weight of the atoms C and H for calculate the molar mass:

Weight of CH4= weight C+ 4 x weight H= 12,01 g/mol +4 x 1,008g/mol=

Weight of CH4 =16, 042 g/mol

1molCH4-----16, 042grams

3,758 mol CH4--X= (3,758 mol CH4 x 16, 042 grams)/1 mol CH4=60,285836 grams

5 0
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