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nikklg [1K]
4 years ago
8

The interactions within the molecule are called? a. intramolecular forces, b. hydrogen bonding,c. intermolecular forces, or d. d

ispersion forces?
Chemistry
1 answer:
Verizon [17]4 years ago
5 0
Intra molecular forces
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Start with moles of ammonia you want to produce, then convert to how many moles of H2 you need to get one mole of ammonia, then convert that to grams :)

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Which group, metals or nonmetals, has the greatest tendency to gain electrons and to become negative ions???
Harlamova29_29 [7]
Nonmetals has the greatest tendency to become negative ions.
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The concentration of Fe2 in a sample is determined by measuring the absorbance of its complex with ferroxine. The sample, measur
sdas [7]

Answer:

  • Absorbance of sample solution = 1.21
  • Absorbance of reagent blank = 0.205

Explanation:

In order to solve this problem we need to keep in mind the <em>Lambert-Beer law</em>, which states:

  • A = ε*b*C

Where ε is the molar absorption coefficient, b is the length of the cuvette, and C is the concentration.

By looking at the equation above we can see that if ε and C are constant; and b is 5 times higher (5.00 cm vs 1.00 cm) then the absorbance will be 5 times higher as well:

  • Absorbance of sample solution = 0.242 * 5 = 1.21
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8 0
3 years ago
Which kelvin temperature is equal to -73 c?
zmey [24]
The conversation rate is c+273=k
8 0
4 years ago
The half-life of a first-order reaction is 13 min. If the initial concentration of reactant is 0.13 M, it takes ________ min for
Zigmanuir [339]

Answer:

Therefore it takes 8.0 mins for it to decrease to 0.085 M

Explanation:

First order reaction: The rate of reaction is proportional to the concentration of reactant of power one is called first order reaction.

A→ product

Let the concentration of A = [A]

\textrm{rate of reaction}=-\frac{d[A]}{dt} =k[A]

k=\frac{2.303}{t} log\frac{[A_0]}{[A]}

[A₀] = initial concentration

[A]= final concentration

t= time

k= rate constant

Half life: Half life is time to reduce the concentration of reactant of its half.

t_{\frac{1}{2} }=\frac{0.693}{k}

Here t_{\frac{1}{2} }=0.13 min

k=\frac{0.693}{t_{\frac{1}{2}} }

\Rightarrow k=\frac{0.693}{13 }

To find the time takes for it to decrease to 0.085 we use the below equation

k=\frac{2.303}{t} log\frac{[A_0]}{[A]}

\Rightarrow t=\frac{2.303}{k} log\frac{[A_0]}{[A]}

Here ,   k=\frac{0.693}{13 },  [A₀] = 0.13 m and [ A] = 0.085 M

t=\frac{2.303}{\frac{0.693}{13} } log(\frac{0.13}{0.085})

\Rightarrow t= 7.97\approx 8.0

Therefore it takes 8.0 mins for it to decrease to 0.085 M

7 0
3 years ago
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