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trapecia [35]
2 years ago
10

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

25.0ºC?
Chemistry
1 answer:
lana [24]2 years ago
5 0

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.

To know more about ideal gas equation, visit the below link:

brainly.com/question/12873752

#SPJ1

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Alika [10]
The claim is that NaCl mixture is a homogeneous mixture.
Homogeneous mixture means that the components of the mixtures cannot be determined or separated by the naked eye. However, these components can be separated using physical means, such as boiling, evaporation and condensation which will be used in this experiment.

First, we need to prepare one molar solution of NaCl. To do so, we will dilute a mass of 58.44 grams (molar mass of NaCl) in 1 liter of water.
By this, we will have NaCl solution.

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..............> observation I

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What we will observe is that when all the water evaporates, we can see white precipitate of NaCl in the bottom of the container. Examining the cold surface placed above the steam, we can see that the water has condensed on this surface.
.........>observation II

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Calculate the moles and grams of solute in each solution. D. 2.0 L of 0.30M Na2SO4. I already have A, B, and C. Thanks!
e-lub [12.9K]
Find the number of moles
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Find the molar mass
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1S   = 32 * 1 = 32 grams
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Find the mass
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85.2 are in a 2 L solution that has a concentration of 0.6 mol/L


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. In a separate experiment, the molar mass of nicotine is found to be somewhere between 150 and 180 g/mol. Calculate the molar m
stealth61 [152]

Answer:

<h2>         162g/mol</h2>

Explanation:

The question is incomplete. The complete question includes the information to find the empirical formula of nicotine:

<em>Nicotine has the formula   </em>C_xH_yN_z<em> . To determine its composition, a sample is burned in excess oxygen, producing the following results:</em>

  • <em>1.0 mol of CO₂</em>
  • <em>0.70 mol of H₂O</em>
  • <em>0.20 mol of NO₂</em>

<em>Assume that all the atoms in nicotine are present as products </em>

<h2>Solution</h2>

To find the empirical formula you need to find the moles of C, H, and N in each of the compound.

  • 1.0 mol of CO₂ has 1.0 mol of C
  • 0.70 mol of H₂O has 1.4 mol of H
  • 0.20 mol of NO₂ has 0.20 mol of N

Thus, the ratio of moles is:

  • C: 1.0
  • H: 1.4
  • N: 0.20

Divide all by the smallest number: 0.20

  • C: 1.0 / 0.20 = 5
  • H: 1.4 / 0.20 = 7
  • N: 0.20 / 0.20 = 1

Hence, the empirical formula is C₅H₇N

Find the mass of 1 mole of units of the empirical formula:

  • C:  5mol  × 12g/mol = 60g
  • H: 7mol × 1g/mol = 7 g
  • N: 1 mol × 14g/mol = 14g

Total mass = 60g + 7g + 14g = 81g

Two moles of units of the empirical formula weighs 2 × 81g = 162g and three units weighs 3 × 81g = 243 g.

Thus, since the molar mass is between 150 and 180 g/mol, the correct molar mass is 162g/mol and the molecular formula is twice the empirical formula: C₁₀H₁₄N₂.

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