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erica [24]
3 years ago
5

A gas is at an original pressure of 412 torr and a temperature of 79°C. What is the new temperature of the gas (in K) if the pre

ssure changes to 341 torr?
Chemistry
1 answer:
olga_2 [115]3 years ago
7 0

Answer:

Whenever you are doing calculations with gases and temperature, you must convert the temperature from C into Kelvin. Then, just plug in your values and do the calculation. So, for the first one,

32C + 273 = 305 K

63C + 273 = 336 K

Then, P1/T1 = P2/T2

575 torr / 305 K = P2 / 336 K

P2 = 633 torr

For the second one:

P1/T1 = P2/T2

575/305 = 5750 / T2

T2 = 3050 K

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6. How many moles of water would require 92.048 kJ of heat to raise its temperature from 34.0 °C to 100.0 °C? (3 marks)​
scoray [572]

Taking into account the definition of calorimetry, 0.0185 moles of water are required.

<h3>Calorimetry</h3>

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

So, the equation that allows to calculate heat exchanges is:

Q = c× m× ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

<h3>Mass of water required</h3>

In this case, you know:

  • Heat= 92.048 kJ
  • Mass of water = ?
  • Initial temperature of water= 34 ºC
  • Final temperature of water= 100 ºC
  • Specific heat of water = 4.186 \frac{J}{gC}

Replacing in the expression to calculate heat exchanges:

92.048 kJ = 4.186 \frac{J}{gC}× m× (100 °C -34 °C)

92.048 kJ = 4.186 \frac{J}{gC}× m× 66 °C

m= 92.048 kJ ÷ (4.186 \frac{J}{gC}× 66 °C)

<u><em>m= 0.333 grams</em></u>

<h3>Moles of water required</h3>

Being the molar mass of water 18 \frac{g}{mole}, that is, the amount of mass that a substance contains in one mole, the moles of water required can be calculated as:

amount of moles=0.333 gramsx\frac{1 mole}{18 grams}

<u><em>amount of moles= 0.0185 moles</em></u>

Finally, 0.0185 moles of water are required.

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2 years ago
H2CO: → H2O + CO2
GaryK [48]

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4 0
3 years ago
How many grams of He are necessary to fill a balloon having a volume of 4.50 × 10^{3} mL to a pressure of 1.14 × 10^{3} torr at
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The mass in grams of He are necessary to fill a balloon having a volume of 4.50×10³ mL to a pressure of 1.14×10³ torr at 25.0ºC is 1.104 grams.

<h3>How do we calculate the grams from moles?</h3>

Mass (W) in grams from moles (n) will be calculated by using the below equation:
n = W/M, where

M = molar mass

Moles of helium gas will be calculated by using the ideal gas equation:

PV = nRT, where

R = universal gas constant = 62.363 L.torr / K.mol

P = pressure of gas = 1.14×10³ torr

V = volume of gas = 4.50×10³ mL = 4.50 L

n = moles of gas = ?

T = temperature = 25.0ºC = 298 K

On putting these values on the ideal gas equation, we get

n = (1140)(4.5) / (62.363)(298) = 5130 / 18584

n = 0.276 moles

Now we convert this moles into mass by using the first equation as:

W = (0.276mol)(4g/mol) = 1.104 g

Hence required mass of helium gas is 1.104 grams.

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