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Vinil7 [7]
3 years ago
13

At sea level water usually boils at 100. °C and 760. mmHg pressure. On Mt Whitney, the pressure is about 560. mmHg. At what temp

erature, in degress Celsius, does water boil on Mt Whitney?

Chemistry
1 answer:
WITCHER [35]3 years ago
7 0

Answer:

  91.7°C

Explanation:

We suppose you have a formula to work from. However, that is not supplied with this problem statement, so we looked one up.

The formula in the attachment is supposed to have good accuracy in the temperature range of interest. It gives vapor pressure of water in kPa, not mmHg, so we needed the conversion for that, too.

560 mmHg corresponds to about 74.66 kPa. The attached "Buck equation" formula is used to find the corresponding temperature. The exponential equation could be solved algebraically using logarithms and the quadratic formula, but we choose to find the solution graphically.

Water boils at about 91.7 °C on Mt. Whitney.

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There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with
grandymaker [24]

\text{Hg} \text{O} and \text{Hg}_{2} \text{O}.

Assuming complete decomposition of both samples,

  • m(\text{Hg}) = m(\text{residure})
  • m(\text{O}) = m(\text{loss})

First compound:

  • m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g
  • m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g

n = m/M; 0.6498 \; g of the first compound would contain

  • n(\text{O atoms}) = 0.048 \; g  / 16 \; g \cdot mol^{-1}= 0.003 \; mol
  • n(\text{Hg atoms}) = 0.6018 \; g  / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol

Oxygen and mercury atoms seemingly exist in the first compound at a 1:1 ratio; thus the empirical formula for this compound would be \text{Hg} \text{O} where the subscript "1" is omitted.

Similarly, for the second compound

  • m(\text{O}) = m(\text{loss}) = 0.016 \; g
  • m(\text{Hg}) = m(\text{residure}) = 0.4172 - 0.016 = 0.4012  \; g

n = m/M; 0.4172 \; g of the first compound would contain

  • n(\text{O atoms}) = 0.016 \; g  / 16 \; g \cdot mol^{-1}= 0.001 \; mol
  • n(\text{Hg atoms}) = 0.4012 \; g  / 200.58 \; g \cdot mol^{-1}= 0.002 \; mol

n(\text{Hg}) : n(\text{O}) \approx  2:1 and therefore the empirical formula

\text{Hg}_{2} \text{O}.

8 0
3 years ago
Need help asap please and thank you!!
ziro4ka [17]

Answer:

Brother/sister It's Layer 'C'

Explanation:

HOPE IT HELPZZ!

8 0
2 years ago
Another gas was used for many years to float transportation balloons.
Alja [10]
Another gas that was used to fill balloons was hydrogen. The reason why hydrogen is not used today is because it use to cause many fires when it was used to fill balloons up.

Hope I helped you! :)
6 0
3 years ago
THIS IS A THREE PART QUESTION IF YOU CAN HELP IT WOULD BE REALLY APPRECIATED SO I DONT FAIL.
Nostrana [21]

Answer:

1. B = 1.13M

2. A. 8.46%

3. D = 0.0199

Explanation:

1. Molarity of a a solution = number of moles of solute/ volume of solution in L

Number of moles of solute = mass of solute/molar mass of solute

Molar mass of NaC₂H₃O₂ = 82 g/mol, mass of NaC₂H₃O₂ = 245 g

Number of moles of NaC₂H₃O₂ = 245 g / 82 g/mol = 2.988 moles

Molarity of solution = 2.988 mols/ 2.65 L = 1.13 M

2. Percentage by mass of a substance = mass of substance /mass of solution × 100%

Mass of 2.65 L of water = density × volume

Density of water = 1 Kg/L = 1000 g/L; volume of waterb= 2.65 L

Mass of water = 1000 g/L × 2.65 L = 2650 g

Mass of solution = mass of water + mass of solute = 245 + 2650 =2895 g

Percentage by mass of NaC₂H₃O₂ = 245/2895 × 100% = 8.46%

3. Mole fraction of NaC₂H₃O₂ = moles of NaC₂H₃O₂/moles of solution

Moles of water = mass /molar mass

Mass of water = 2650 g; molar mass of water = 18 g/mol

Moles of water = 2650 g / 18 g/mol = 147.222 moles

Moles of solution = moles of solute + moles of water = 147.222 + 2.988 = 150.21

Moles of NaC₂H₃O₂ =2.988 moles

Moles fraction of NaC₂H₃O₂ =2.988/150.21 = 0.0199

6 0
3 years ago
When 8.21 l of c3h8 (g) burn in oxygen, how many liters of oxygen are consumed? all gas volumes are measured at the same tempera
just olya [345]
<span>8.21 L of C3H8(g) Lets take c as the molar volume at that temperature.
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  C3H8 (g) + 5O2 (g) --> 3CO2 + 4H2O + Q
  8.21 L <><> x L
    x = (8.21 * 5c)/c = 8.21 * 5 = 41.05 L O2 consumed for a 100% yield.</span>
5 0
3 years ago
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