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dolphi86 [110]
3 years ago
6

What is the molarity of the following?

Chemistry
1 answer:
pishuonlain [190]3 years ago
7 0
Molarity = number mole of substances/ Volume solution (L)

To find number of mole we need to find Molar mass , at first.

1) M(BaBr2)= M(Ba) +2 M(Br) = 137.3 + 2*80.0= 197.3 g/mol
23g BaBr2 *(1mol BaBr2/197.3 g BaBr2) = 0.12 mol BaBr2
Molarity (BaBr2 solution) =0.12 mol/2.3 L≈0.052 mol/L ( BaBr2 solution)
Molarity of BaBr2 solution = 0.052 M


2) M(CaS)= M(Ca)+M(S)= 40.1+32.1=72.2 g/mol
15 g CaS*(1mol CaS/72.2 g CaS)= 0.21 mol CaS
Molarity (CaS solution) = 0.21 mol/2 L≈ 0.11 mol/L (CaS solution)
Molarity of CaS solution = 0.11 M
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A beam of light passes though a liquid in a test tube without scattering. Which type of mixture is most likely in the test tube?
butalik [34]
The answer would be letter C - solution.

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you just got home from a run on a hot Atlanta afternoon. you grab a 1.00-liter bottle of water and drink three-quarters of it in
nikitadnepr [17]

Answer:

41.67 mol

Explanation:

1 Litre of water = 1000g

Mole = mass / molar mass

Mass of 1 L of water = 1000 g

Molar mass of water (H2O) :

(H = 1, O = 16)

H2O = (1 * 2) + 16 = (2 + 16) = 18g/mol

Amount of water consumed = (3/4) of 1 litre

= (3/4) * 1000g

= 750g

Therefore mass of water consumed = 750g

Mole = 750g / 18g/mol

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2 years ago
When light is shown on a mixture of chlorine and chloromethane, carbon tetrachloride is one of the components of the final react
velikii [3]

A free-radical substitution reaction is likely to be responsible for the observations. The reaction mechanism of a reaction like this can be grouped into three phases:

  • Initiation; the "light" on the mixture deliver sufficient amount of energy such that the halogen molecules undergo homologous fission. It typically takes ultraviolet radiation to initiate fissions of the bonds.  
  • Propagation; free radicals react with molecules to produce new free radicals and molecules.
  • Termination; two free radicals combine and form covalent bonds to produce stable molecules. Note that it is possible for two carbon-containing free-radicals to combine, leading to the production of trace amounts of long carbon chains in the product.

Initiation

\text{Cl}-\text{Cl} \stackrel{\text{UV}}{\to} \text{Cl}\bullet + \bullet\text{Cl}

where the big black dot indicates unpaired electrons attached to the atom.

Propagation

\text{CH}_3\text{Cl}+ \text{Cl}\bullet \to \bullet\text{CH}_2\text{Cl} + \text{HCl}

\bullet\text{CH}_2\text{Cl} + \text{Cl}_2 \to \text{CH}_2\text{Cl}_2 + \text{Cl}\bullet

\text{CH}_2\text{Cl}_2 + \text{Cl}\bullet \to \bullet\text{CHCl}_2 + \text{HCl}

\bullet\text{CHCl}_2+ \text{Cl}_2 \to \text{CHCl}_3 + \bullet \text{Cl}

\text{CHCl}_3 + \text{Cl}\bullet \to \bullet\text{CCl}_3 + \text{HCl}

\bullet\text{CCl}_3 + \text{Cl}_2 \to \text{CCl}_4 + \text{Cl}\bullet

Termination

\text{Cl}\bullet + \bullet\text{Cl} \to \text{Cl}-\text{Cl}

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Answer:

a, b, c, d

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Rutherford’ atomic model is based on the gold foil experiment. In this experiment, beam of alpha rays was bombarded on thin gold foil. He observed that:

Most of the alpha particles passed through thin foil without any deflection.

Few alpha particles deflected by an angle of 90o.

Based on observation, Rutherford concluded that majority of the space inside the atom is empty.

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The mass of an atom is concentrated at the nucleus

Therefore, the correct options are:

a, b, c, d

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