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melisa1 [442]
3 years ago
9

A 3.0-L balloon containing helium gas, He, at 17 °C, was taken outdoors where the temperature was close to 0 °C. What happened t

o the
volume of the balloon?
The gas volume increased.
The gas volume decreased.
The gas volume remains constant.
O With the provided information, gas volume cannot be modeled.
Chemistry
1 answer:
kvv77 [185]3 years ago
7 0

Answer:

The gas volume remains constant

Explanation:

The volume stayed constant however the balloon may seem to have shriveled or deflated upon walking outside due to the helium condensing in  the cooler climate.

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Which location will experience the greatest amount of weathering
Ivanshal [37]

Answer:

Explanation:

Weathering occurs fastest in hot, wet climates. It occurs very slowly in hot and dry climates.

4 0
3 years ago
a weather balloon contains 8.80 moles of helium at a ppressure of 0.992 atm and a temperature os 25 C at ground level. What is t
djverab [1.8K]
We can use the ideal gas law equation to find the volume of the balloon.
PV = nRT 
where 
P - pressure - 0.992 atm x 101 325 Pa/atm = 100 514 Pa
V - volume 
n - number of moles - 8.80 mol 
R - universal gas constant  - 8.314 Jmol⁻¹K⁻¹
T - temperature in kelvin - 25 °C + 273 = 298 K
Substituting these values in the equation 
100 514 Pa x V = 8.80 mol x  8.314 Jmol⁻¹K⁻¹ x 298 K
V = 217 L
volume of balloon is 217 L
7 0
3 years ago
What is the volume of 12.0 grams of oxygen gas at STP<br><br> Atomic mass:O = 15.99 grams/moles
strojnjashka [21]

Answer:

16.82 L.

Explanation:

  • We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm (P = 1.0 atm, STP conditions).

V is the volume of the gas in L (V = ??? L).

n is the no. of moles of the gas in mol (n = mass/molar mass = (12.0 g)/(15.99 g/mol) = 0.7505 mol).

R is the general gas constant (R = 0.0821 L.atm/mol.K),

T is the temperature of the gas in K (T = 0.0°C + 273 = 273.0 K, STP conditions).

<em>∴ V = nRT/P</em> = (0.7505 mol)(0.0821 L.atm/mol.K)(273.0 K)/(1.0 atm) = <em>16.82 L.</em>

6 0
4 years ago
In the laboratory a student determines the specific heat of a metal. She heats 19.6 grams of zinc to 98.37°C and then drops it i
fgiga [73]

Answer:

The specific heat of zinc is 0.375 J/g°C

Explanation:

<u>Step 1:</u> Data given

Mass of zinc = 19.6 grams

Mass of water = 82.9 grams

Initial temperature of zinc = 98.37 °C

Initial temperature of water = 24.16 °C

Final temperature of water (and zinc) = 25.70 °C

Specific heat of water = 4.184 J/g°C

<u>Step 2:</u> Calculate Specific heat of zinc

Heat lost by zinc = heat won by water

Q=m*c*ΔT

Qzinc = -Qwater

m(zinc)*C(zinc)*ΔT(zinc) = -m(water)*C(water)*ΔT(water)

⇒ with mass of zinc = 19.6 grams

⇒ with C(zinc) = TO BE DETERMINED

⇒ with ΔT(zinc) = T2 -T1 = 25.70 - 98.37 = -72.67°C

⇒ with mass of water = 82.9 grams

⇒ with C(water) = 4.184 J/g°C

⇒ with ΔT(water) = T2 - T1 = 25.70 - 24.16 = 1.54

Qzinc = -Qwater

19.6g* C(zinc) * (-72.67°C) = - 82.9g* 4.184 J/g°C * 1.54 °C

-1424.332*C(zinc) = -534.155

C(zinc) = 0.375 J/g°C

The specific heat of zinc is 0.375 J/g°C

7 0
3 years ago
At 450 mm Hg a gas has a volume of 760 L, what is its volume at standard pressure​
KatRina [158]

Answer:

450.0 L.

Explanation:

  • We can use the general law of ideal gas: <em>PV = nRT.</em>

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

  • If n and T are constant, and have different values of P and V:

<em>(P₁V₁) = (P₂V₂).</em>

<em></em>

V₁ = 760.0 L, P₁ = 450.0 mm Hg,

V₂ = ??? L, ​P₂ = 760.0 mm Hg (standard pressure = 1.0 atm = 760 mm Hg).

∴ V₂ = (P₁V₁)/(P₂) = (760.0 L)(450.0 mm Hg)/(760.0 mm Hg) = 450.0 L.

5 0
4 years ago
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