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MA_775_DIABLO [31]
3 years ago
14

Which is a radical? A. OB. HC. OHD. H2O​

Chemistry
1 answer:
Gnesinka [82]3 years ago
8 0

Answer: C. OH

Explanation: A radical is defined as a molecule with unpaired electrons; thus your answer is OH. However, if you want a more general definition (an atom or molecule with unpaired electrons), then OH, O and H are all radicals. Only H2O is a closed-shell species. and is therefore not a radical.

Hope this helps! :)

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Answer:

D.

atomic modifier

Explanation:

. Identify at least three reasons the Articles of Confederations failed as a governing document. In your opinion, evaluate which defect was most debilitating, using evidence and your knowledge of American government to justify your position. [5]

6 0
4 years ago
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Suppose now that you wanted to determine the density of a small crystal to confirm that it is phosphorus. From the literature, y
denis23 [38]

Answer:

Volume of CHCl_3   = 15.31 mL

Volume of CHBr_3 = 4.69 mL

Explanation:

Given that:

the density of the mixture = 1.82 g/mL

From the density of the pure samples

The density of CHCl_3 = 1.492 g/mL

The density of CHBr_3 = 2.890 g/mL

The total volume of the liquid mixture = 20.0 mL

Suppose the volume of  CHCl_3 = P ml

and the volume of CHBr_3 = Q ml

the sum of their volumes should be equal to the total volume of the mixture

P \ ml + Q \ ml = 20 ml  ----- (1)

However, we know that Density = mass/volume

∴ mass = density × volume

The equation can now be expressed as:

\mathtt{(Density \ of  \ CHCl_3 \times Vol. \ of  \ CHCl_3 ) + (Density  \  of \  CHBr_3  \times \ volume \ of \ CHBr_3)} = \mathtt{ (Density  \ of \ mixture \times volume \ of \ the \ mixture)}

1.492 g/mL × P mL + 2.890 g/mL × Q mL = 1.82 g/mL × 20 mL  ---- (2)

From equation (1) ;

let Q = 20 - P

The replace the value of P into equation (2)

1.492 g/mL × P mL + 2.890g/mL × (20 - P) mL = 1.82 g/mL × 20 mL

1.492 P g + 57.8g - 2.890 P g =  36.4g

1.492 P g - 2.890 P g = 36.4g - 57.8g

-1.398 P g = -21.4g

P = -21.4g/-1.398g

P = 15.31 mL

Q = 20 - P

Q = (20 - 15.31) mL

Q = 4.69 mL

∴

Volume of CHCl_3   = 15.31 mL

Volume of CHBr_3 = 4.69 mL

6 0
3 years ago
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Anna11 [10]

Answer:

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5 0
3 years ago
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A. Based on the activation energies and frequency factors, rank the following reactions from fastest to slowest reaction rate, a
deff fn [24]

Answer:

A) E_{a} = 350KJ/mol, E_{a} = 50KJ/mol, E_{a} = 50KJ/mol

     A = 1.5×10^{-7}s^{-1}, A = 1.9×10^{-7} s^{-1}, A=1.5×10^{-7} s^{-1}

B) 4.469

Explanation:

From Arrhenius equation

      K=Ae^{\frac{E_{a} }{RT} }

where; K = Rate of constant

            A = Pre exponetial factor

            E_{a} = Activation Energy

             R = Universal constant

             T = Temperature in Kelvin

Given parameters:

E_{a} =165KJ/mol

T_{1}=505K

T_{2}=525K

R=8.314JK^{-1}mol^{-1}

taking logarithm on both sides of the equation we have;

InK=InA-\frac{E_{a} }{RT}

since we have the rate of two different temperature the equation can be derived as:

In(\frac{K_{2} }{K_{1} } )=\frac{E_{a} }{R}(\frac{1}{T_{1} } -\frac{1}{T_{2} } )

In(\frac{K_{2} }{K_{1} } )=\frac{165000J/mol}{8.314JK^{-1}mol^{-1}  }.(\frac{1}{505} -\frac{1}{525} )

In(\frac{K_{2} }{K_{1} } )= 19846.04×7.544×10^{-5} = 1.497

\frac{K_{2} }{K_{1} } =e^{1.497} = 4.469

 

6 0
3 years ago
When doing a titration what is the equivalency point of the neutralization reaction?
KengaRu [80]
The equivalency point is at the point of the titration where the amount of titrant added neutralize the solution. When it’s a strong acid strong base titration, the equivalence point will be 7. When it is a weak acid strong base, the equivalence point it more basic (the exact number depends on what acid and base you use). And when it is a strong acid weak base, the equivalence number is more acid (the exact number depends on what acid and base you use). Hope this helps!
3 0
3 years ago
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