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NikAS [45]
1 year ago
15

What volume did a helium-filled balloon have at 22.5 c and 1.95 atm if it’s new volume was 56.4 mL at 3.69 atm and 11.9c

Chemistry
2 answers:
Stells [14]1 year ago
8 0

Answer:

110.69 ml

Explanation:

Not sure why my answer was removed:

Use the general rule for gases:

P1 V1 / T1    =   P2 V2 / T2      Looking for V1     T must be in Kelvin

re-arrange to :

V1 =   P2 V 2  * T1 / (T2 * P1)      <==== now sub in the values

V1 = 3.69 * 56.4 * (22.5 + 273.15) / [(11.9 + 273.15) * 1.95]

V1 = 110.69 ml

Veseljchak [2.6K]1 year ago
6 0

This is an exercise in the general or combined gas law.

To start solving this exercise, we obtain the following data:

<h3>Data:</h3>
  • T₁ = 22.5 °C + 273 = 295.5 K
  • P₁ = 1.95 atm
  • V₁ = ¿?
  • P₂ = 3.69 atm
  • T₂ = 11.9 °C + 273 =  284.9 k
  • V₂= 56.4 ml

We use the following formula:

P₁V₁T₂ = P₂V₂T₁ ⇒ General formula

Where

  • P₁ = Initial pressure
  • V₁ = Initial volume
  • T₂ = Initial temperature
  • P₂ = Final pressure
  • V₂ = final volume
  • T₁ = Initial temperature

We clear the formula for the initial volume:

\boldsymbol{\sf{V_{1}=\dfrac{P_{2}V_{2}T_{1}}{P_{1}T_{2}}  } }

We substitute our data into the formula to solve:

\boldsymbol{\sf{V_{1}=\dfrac{(3.69 \not{atm})(56.4 \ ml)(295.5 \not{k})}{(1.95 \not{atm})(284.9\not{k})}  }}

\boldsymbol{\sf{V_{1}=\dfrac{61498.278}{555.555} \ lm }}

\boxed{\boldsymbol{\sf{V_{1}=110.697 \ lm }}}

The helium-filled balloon has a volume of <u>110.697 ml.</u>

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A 60.0 g block of iron that has an initial temperature of 250. °C and 60.0 g bloc of gold that has an initial temperature of 45.
Maslowich

Answer:

The final temperature at the equilibrium is 204.6 °C

Explanation:

Step 1: Data given

Mass of iron = 60.0 grams

Initial temperature = 250 °C

Mass of gold = 60.0 grams

Initial temperature of gold = 45.0 °C

The specific heat capacity of iron = 0.449 J/g•°C

The specific heat capacity of gold = 0.128 J/g•°C.

Step 2: Calculate the final temperature at the equilibrium

Heat lost = Heat gained

Qlost = -Qgained

Qiron = -Qgold

Q=m*c*ΔT

m(iron) * c(iron) *ΔT(iron) = -m(gold) * c(gold) *ΔT(gold)

⇒with m(iron) = the mass of iron = 60.0 grams

⇒with c(iron) = the specific heat of iron = 0.449 J/g°C

⇒with ΔT(iron)= the change of temperature of iron = T2 - T1 = T2 - 250.0°C

⇒with m(gold) = the mass of gold= 60.0 grams

⇒with c(gold) = the specific heat of gold = 0.128 J/g°C

⇒with ΔT(gold) = the change of temperature of gold = T2 - 45.0 °C

60.0 *0.449 * (T2 - 250.0) = -60.0 * 0.128 * (T2 - 45.0 )

26.94 * (T2 - 250.0) = -7.68 * (T2 - 45.0)

26.94T2 - 6735 = -7.68T2 + 345.6

34.62T2 = 7080.6

T2 = 204.5 °C

The final temperature at the equilibrium is 204.6 °C

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