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yulyashka [42]
1 year ago
15

Under identical conditions of temperature, number of moles, and pressure, which of the following gases has the highest density:

Ne, CO2, or Cl2?
Chemistry
2 answers:
sergey [27]1 year ago
6 0
<span>The higher the molar mass is of the gas, the greater the density.

Cl2 is the answer</span>
Ulleksa [173]1 year ago
6 0

Answer: Option (c) is the correct answer.

Explanation:

As density is mass divided by volume.

Mathematically,      Density = \frac{mass}{volume}

So, more is the mass of a substance more will be its density but if volume is low then density will also have a low value.

Molar mass of neon is 20 g/mol.

Molar mass of CO_{2} is 44 g/mol.

Molar mass of Cl_{2} is 70.9 g/mol.

Since, Cl_{2} molecule has the highest mass. Therefore, its density will also be high.

Thus, we can conclude that out of the given options Cl_{2} gas has the highest density.

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A sample of neon gas at a pressure of 1.08 atm fills a flask with a volume of 250 mL at a temperature of 24.0 °C. If the gas is
musickatia [10]

Answer:

124.91mL

Explanation:

Given parameters:

P₁  = 1.08atm

V₁  = 250mL

T₁  = 24°C

P₂  = 2.25atm

T₂  = 37.2°C

V₂  = ?

Solution:

To solve this problem, we are going to apply the combined gas law;

              \frac{P_{1} V_{1} }{T_{1} }   =  \frac{P_{2} V_{2} }{T_{2} }

P, V and T represents pressure, volume and temperature

1 and 2 delineates initial and final states

Convert the temperature to kelvin;

        T₁  = 24°C,  T₁   = 24 + 273 = 297K

        T₂  = 37.2°C , T₂  = 37.2 + 273  = 310.2K

Input the variables and solve for V₂

        \frac{1.08 x 250}{298} = \frac{2.25 x V_{2} }{310.2}

           V₂ = 124.91mL

6 0
2 years ago
I NEED HELP ASAP, WILL MARK BRAINLEST!
Andre45 [30]

Answer:

1. 90%

2. 217.4 g O₂

3. 95.0%

4. Trial 2 ratios

Explanation:

Original: SiCl₄ + O₂ → SiO₂ + Cl₂

Balanced: SiCl₄ + O₂ → SiO₂ + 2Cl₂

Trial        SiCl₄                   O₂                    SiO₂

 1           120 g                  240 g              38.2 g

 2           75 g                   50 g                25.2 g

<u>Percentage yield for trial 1</u>

We need to get actual yield (38.2 g) and theoretical yield, in grams.

Mass to moles:

 molar mass SiCl₄: 28.09 + 4(35.45) = 169.9 g/mol

 120 g SiCl₄ x 1 mol/169.9 g = .706 mol SiCl₄

Moles to moles:

 For each mole SiCl₄, we have one mol SiO₂ based on the balanced rxn.

 .706 mol SiCl₄ = .706 mol SiO₂

Moles to mass:

 molar mass SiO₂: 28.09 + 2(16.00) = 60.09 g/mol

 .706 mol SiO₂ x 60.09g/mol = 42.44 g SiO₂

Theoretical yield:

 actual/theoretical x 100

 38.2 / 42.44 = .900 = <u>90.0% yield</u>

<u>Leftover reactant for trial 1</u>

We know oxygen is the excess reactant.

Mass to moles:

 molar mass O₂ = 32.00 g/mol

 240 g O₂ x 1 mol/32.00 g = 7.5 mol O₂

We used .706 mol SiO₂, so we also used .706 mol O₂.

 7.5 - .706 = 6.8 moles left over

Moles to mass:

 6.8 mol O₂ x 32.00g/mol =<u> 217.4 g O₂</u>

<u />

<u>Percentage yield for trial 2</u>

Mass to moles:

 molar mass SiCl₄: 169.9 g/mol

 75 g SiCl₄ x 1 mol/169.9 g = .441 mol SiCl₄

Moles to moles:

 For each mole SiCl₄, we have one mol SiO₂ based on the balanced rxn.

 .441 mol SiCl₄ = .441 mol SiO₂

Moles to mass:

 molar mass SiO₂: 60.09 g/mol

 .441 mol SiO₂ x 60.09g/mol = 26.5 g SiO₂

Theoretical yield:

 actual/theoretical x 100

 25.2 / 26.5 = .950 = <u>95.0% yield</u>

Because the percentage yield of trial 2 is higher than that of trial 1, we know that the ratio of reactants in trial 2 is more efficient! We got a result closer to our theoretical yield.

6 0
2 years ago
Arkeisha and Rodney are having a discussion about the correct use of the term "battery acid." Arkeisha insists that an alkaline
mote1985 [20]
Arkeisha is correct because the fluid in an alkaline battery has a ph between 7.1 and 14.0
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The health codes cover all issues, except:
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Answer:

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2 years ago
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A 0.307-g sample of an unknown triprotic acid is titrated to the third equivalence point using 35.2 ml of 0.106 m naoh. calculat
Jet001 [13]
Triprotic acid is a class of Arrhenius acids that are capable of donating three protons per molecule when dissociating in aqueous solutions.  So the chemical reaction as described in the question, at the third equivalence point, can be show as: H3R + 3NaOH ⇒ Na3R + 3H2O, where R is the counter ion of the triprotic acid. Therefore, the ratio between the reacted acid and base at the third equivalence point is 1:3. 
The moles of NaOH is 0.106M*0.0352L = 0.003731 mole.  So the moles of H3R is 0.003731mole/3=0.001244mole.
The molar mass of the acid can be calculated: 0.307g/0.001244mole=247 g/mol.
6 0
2 years ago
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