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aliina [53]
3 years ago
7

Explain why the air inside of the balloon has more pressure than the air outside.

Chemistry
1 answer:
julia-pushkina [17]3 years ago
6 0

Answer:

its more condensed

Explanation:

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What was the initial temperature of a gas contained in a 2.0-L bottle at a pressure of 1.00 atm if it was placed under vacuum an
lesya692 [45]
V1/T1 = V2/T2 

Substitute the value use ratio and proportion. Use calculator. 

V1 = (V2 x T1) / T2

1 is initial, 2 is final
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3 years ago
How to find change in enthalpy
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∆H = m x s x ∆T, where m is the mass of the reactants, s is the specific heat of the product, and ∆T is the change in temperature from the reaction.
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A sample of hydrogen gas collected over water occupied 30.0 mL at 24 °C on a day when the atmospheric pressure was 736 Torr. Wha
givi [52]

Answer:0.026ml

Explanation:

Details are found in the image attached. We must subtract the saturated vapour pressure of hydrogen gas at the given temperature from the total pressure of the hydrogen gas collected over water to obtain the actual pressure of hydrogen gas and substitute the value obtained into the general gas equation. The dry hydrogen gas has no saturated vapour pressure hence the value is substituted as given. All temperatures must be converted to Kelvin before substitution.

4 0
3 years ago
2. A quantity of 1.922g of methanol (CH3OH) was burned in a constant-volume
Cerrena [4.2K]
Mass of methanol (CH3OH) = 1.922 g
Change in Temperature (t) = 4.20°C
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The heat absorbed by the bomb and water is equal to the product of the heat capacity and the temperature change.
Let’s assume that no heat is lost to the surroundings. First, let’s calculate the heat changes in the calorimeter. This is calculated using the formula shown below:
qcal = Ccalt
Where, qcal = heat of reaction
Ccal = heat capacity of calorimeter
t = change in temperature of the sample
Now, let’s calculate qcal:
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= 43.68 kJ
Always qsys = qcal + qrxn = 0,
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The heat change of the reaction is - 43.68 kJ which is the heat released by the combustion of 1.922 g of CH3OH. Therefore, the conversion factor is:
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A speed time graph shows that a car moves at 20 m/s for 15s. The car's speed steadily decreases until it comes to a stop at 40s.
ladessa [460]
0s to 15s: constant speed/zero acceleration
15s to 40s: constant gradient, therefore constant deceleration
4 0
3 years ago
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