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bixtya [17]
2 years ago
9

An unknown gas effuses at a rate of 2.0 times the rate of Cl₂. What is the molar mass of the unknown gas?

Chemistry
1 answer:
Ratling [72]2 years ago
3 0

The molar mass of the unknown gas is determined as 17.75 g.

<h3>Rate of gas diffusion</h3>

The rate of gas diffusion is given by Graham's law.

R₁√M₁ = R₂√M₂

where;

  • R₁ is the rate of diffusion of the unknown gas
  • M₁ is the molecular mass of the unknown gas
  • R₂ is rate of diffusion of chlorine gas
  • M₂ is the molecular mass of chlorine gas

\frac{R_1}{R_2} = \sqrt{\frac{M_2}{M_1} } \\\\\frac{2R_2}{R_2} = \sqrt{\frac{71}{M_1} }\\\\2 = \sqrt{\frac{71}{M_1} }\\\\2^2 = \frac{71}{M_1} \\\\M_1 = \frac{71}{4} \\\\M_1 = 17.75 \ g

Learn more about rate of diffusion here:  brainly.com/question/11197959

#SPJ1

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Hydrazine (N2H4) is a fuel used by some spacecraft. It is normally oxidized by N2O4 according to the following equation: N2H4(l)
vitfil [10]

Answer:

The enthalpy of the reaction is coming out to be -380.16 kJ.

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as \Delta H

The equation used to calculate enthalpy change is of a reaction is:  

\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

For the given chemical reaction:

N_2H_4(l)+N_2O_4(g)\rightarrow 2N_2O(g)+2H_2O(g)

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(2 mol\times \Delta H_f_{(N_2O)})+(2 mol\times\Delta H_f_{(H_2O)} )]-[(1 mol\times \Delta H_f_{(N_2H_4)})+(1 mol\times \Delta H_f_{(N_2O_4)})]

We are given:

\Delta H_f_{(N_2O)}=81.6 kJ/mol\\\Delta H_f_{(H_2O)}=-241.8 kJ/mol\\\Delta H_f_{(N_2H_4)}= 50.6 kJ/mol\\\Delta H_f_{(N_2O_4)}=9.16 kJ/mo

Putting values in above equation, we get:

\Delta H_{rxn}=[(2 mol\times 81.6 kJ/mol)+2 mol\times -241.8 kJ/mol)]-[(1 mol\times (50.6 kJ/mol))+(1 mol\times (9.16))]\\\\\Delta H_{rxn}=-380.16 kJ

Hence, the enthalpy of the reaction is coming out to be -380.16 kJ.

6 0
3 years ago
Identify the hybridization of each carbon atom for the molecule above
ipn [44]

Carbons starting from the left end:

  1. sp²
  2. sp²
  3. sp²
  4. sp
  5. sp

Refer to the sketch attached.

<h3>Explanation</h3>

The hybridization of a carbon atom depends on the number of electron domains that it has.

Each chemical bond counts as one single electron domain. This is the case for all chemical bonds: single, double, or triple. Each lone pair also counts as one electron domain. However, lone pairs are seldom seen on carbon atoms.

Each carbon atom has four valence electrons. It can form up to four chemical bonds. As a result, a carbon atom can have up to four electron domains. It has a minimum of two electron domains, with either two double bonds or one single bond and one triple bond.

  • A carbon atom with four electron domains is sp³ hybridized;
  • A carbon atom with three electron domains is sp² hybridized;
  • A carbon atom with two electron domains is sp hybridized.

Starting from the left end (H₂C=CH-) of the molecule:

  • The first carbon has three electron domains: two C-H single bonds and one C=C double bond; It is sp² hybridized.
  • The second carbon has three electron domains: one C-H single bond, one C-C single bond, and one C=C double bond; it is sp² hybridized.
  • The third carbon has three electron domains: two C-C single bonds and one C=O double bond; it is sp² hybridized.
  • The fourth carbon has two electron domains: one C-C single bond and one C≡C triple bond; it is sp hybridized.
  • The fifth carbon has two electron domains: one C-H single bond and one C≡C triple bond; it is sp hybridized.

8 0
3 years ago
In the reaction H2SO4 + 2 NaOH -&gt; Na2SO4 + 2H2O, an equivalence point occurs when 29.43 mL of 0.1973 M NaOH is added to a 32.
Tamiku [17]
            moles NaOH = c · V = 0.1973 mmol/mL · 29.43 mL = 5.806539 mmol
            moles H2SO4 = 5.806539 mmol NaOH · 1 mmol H2SO4 / 2 mmol NaOH = 2.9032695 mmol
Hence
            [H2SO4]= n/V = 2.9032695 mmol / 32.42 mL = 0.08955 M
The answer to this question is  [H2SO4] = 0.08955 M

6 0
3 years ago
If you burn 55.6 g of hydrogen and produce 497 g of water, how much oxygen reacted?
tester [92]

Answer:

441.28 g Oxygen

Explanation:

  • The combustion of hydrogen gives water as the product.
  • The equation for the reaction is;

2H₂(g) + O₂(g) → 2H₂O(l)

Mass of hydrogen = 55.6 g

Number of moles of hydrogen

Moles = Mass/Molar mass

          = 55.6 g ÷ 2.016 g/mol

          = 27.8 moles

The mole ratio of Hydrogen to Oxygen is 2:1

Therefore;

Number of moles of oxygen = 27.5794 moles ÷ 2

                                               = 13.790 moles

Mass of oxygen gas will therefore be;

Mass = Number of moles × Molar mass

Molar mass of oxygen gas is 32 g/mol

Mass = 13.790 moles × 32 g/mol

<h3>          = 441.28 g</h3><h3>Alternatively:</h3>

Mass of hydrogen + mass of oxygen = Mass of water

Therefore;

Mass of oxygen = Mass of water - mass of hydrogen

                          = 497 g - 55.6 g

<h3>                           = 441.4 g </h3>
6 0
3 years ago
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denpristay [2]
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5 0
3 years ago
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